Quantitative chemistry

Question 1

What is the law of conservation of mass?

Answer 1

The idea that no atoms are lost or made during a chemical reaction, therefore, the mass of the products equals the mass of the reactants and the number of atoms in each element is the same.

Question 2

What is a reactant?

Answer 2

The starting substances in a reaction.

Question 3

What is a product?

Answer 3

The chemicals formed in a reactant.

Question 4

Question:

92g of sodium reacts with 32g of oxygen. Calculate the mass of sodium oxide produced.

4Na + O₂ ⭢ 2Na₂O

Answer 4

92 + 32 = 124g

Total mass of product(s) = total mass of reactant(s)

Question 5

Question:

A mass of calcium carbonate reacted to produce 112g of calcium oxide and 88g of carbon dioxide. Calculate the mass of calcium carbonate that reacted.

CaCO₃ ⭢ CaO + CO₂

Answer 5

112 + 88 = 200g

Total mass of reactant(s) = total mass of product(s)

Question 6

Question:

A mass of magnesium oxide reacts with 73g of hydrogen chloride to produce 95g of magnesium chloride and 18g of water. Calculate the mass of magnesium oxide in the reaction.

MgO + 2HCl ⭢ MgCl₂ + H₂O

Answer 6

95 + 18 = 113g
113 - 73 = 40g

Total mass of product(s) = total mass of reactant(s)

Question 7

How [when are] are ionic compounds formed?

Answer 7

When metals react with non-metals.

Question 8

What are ions?

Answer 8

Atoms of an element with an overall charge that are formed when atoms gain or lose electrons.

Question 9

What ions do metals / [most] non-metals form?

Answer 9

• Metals form positive ions.
• Most non-metals form negative ions.

Question 10

What often correlates with the charge of an ion? What does not follow this pattern? Why?

Answer 10

The group number of the metals;
* transition metals as they can have [form] several different [ions] charges
* non-metal ionic compounds as they consist of several nonmetal atoms.

Question 11

What are the charges of the non-metal ionic compounds?

Answer 11

Hydroxide: OH⁻ (1-)
Nitrate: NO₃⁻ (1-)
Sulfate: SO₄²⁻ (2-)
Carbonate: CO₃²⁻ (2-)
Ammonium: NH₄⁺ (1+)

Question 12

Question:

Find the formula, using the charges, for sodium chloride.

Answer 12

Na⁺ + Cl⁻ ⭢ NaCl

Question 13

Question:

Find the formula, using the charges, for sodium oxide.

Answer 13

Na⁺ + O²⁻ ⤚ ^{(Na × 2)} ⭢ Na₂O

Question 14

Question:

Find the formula, using the charges, for magnesium iodide.

Answer 14

Mg²⁺ + I⁻ ⤚ ^{(I × 2)} ⭢ MgI₂

Question 15

Question:

Find the formula, using the charges, for lithium carbonate.

Answer 15

Li⁺ + CO₃²⁻ ⤚ ^{(Li × 2)} ⭢ Li₂CO₃

Question 16

Question:

Find the formula, using the charges, for calcium hydroxide.

Answer 16

Ca²⁺ + OH⁻ ⤚ ^{(OH × 2)} ⭢ Ca(OH)₂

Question 17

Question:

Find the formula, using the charges, for magnesium nitrate.

Answer 17

Mg²⁺ + NO₃⁻ ⤚ ^{(NO₃ × 2)} ⭢ Mg(NO₃)₂

Question 18

Question:

Balance the equation:

Calcium oxide + hydrochloric acid ⭢ Calcium chloride + water
CaO + HCl ⭢ CaCl₂ + H₂O

Answer 18

CaO + 2HCl ⭢ CaCl₂ + H₂O

Question 19

Question:

Balance the equation:

Iron oxide + carbon monoxide ⭢ Iron + carbon dioxide
Fe₂O₃ + CO ⭢ Fe + CO₂

Answer 19

Fe₂O₃ + 3CO ⭢ 2Fe + 3CO₂

Question 20

What is the Relative Atomic Mass (Ar)?

Answer 20

The average mass of the atoms of an element, taking into account naturally occurring isotopes, compared with carbon-12 (which is given a mass of exactly 12).

Question 21

What is the Relative Formula Mass (Mr) of a compound?

Answer 21

The sum of the relative atomic masses (Ar) of the atoms in the numbers shown in the formula.

Question 22

What are two things to watch out for when calculating the Relative Formula Mass (Mr)?

Answer 22

1. They have no units
2. Do not take into account coefficients (big numbers) when calculating the Mr.

Question 23

Question:

Calculate the Mr of CaSO₄

(Ar of Ca = 40), (Ar of S = 32), (Ar of O = 16)

Answer 23

Mr = 40 + 32 + (16 × 4) = 136

Question 24

Question:

Calculate the Mr of Mg(OH)₂

(Ar of Mg = 24), (Ar of O = 16), (Ar of H = 1)

Answer 24

Mr = 24 + (16 × 2) + (1 × 2) = 58

Question 25

What is the equation for calculating the percentage composition (percentage by mass)?

Answer 25

% composition: 100 (total Ar of element ÷ total Mr of compound)

Question 26

Question:

Calculate the percentage by mass of calcium (Ca) in calcium chloride (CaCl₂).

Answer 26

Ar of Ca = 40
Mr of CaCl₂ = 111
Percentage by mass of Ca = (40 ÷ 111) × 100 = (0.3603) × 100 = 36%

% composition: 100 (total Ar of element ÷ total Mr of compound)

Question 27

Question:

Calculate the percentage by mass of hydrogen (H) in calcium chloride (CH₄).

Answer 27

Ar of H = 1
Mr of CH₄ = 16
Percentage by mass of H = ((4 × 1) ÷ 16) × 100 = (0.25) × 100 = 25%

% composition: 100 (total Ar of element ÷ total Mr of compound)

Question 28

Question:

Calculate the percentage by mass of sodium (Na) in sodium sulphate (Na₂SO₄).

Answer 28

Mr of Na₂SO₄ = (23 × 2) + 32 + (16 × 4) = 142
Ar of Na = 23
Percentage by mass of Na = ((23 × 2) ÷ 142) × 100 = (0.3239) × 100 = 32.4%

% composition: 100 (total Ar of element ÷ total Mr of compound)

Question 29

What is the equation for calculating the number of moles of an element?

[Higher tier]

Answer 29

Number of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 30

Question:

You are given a sample of magnesium with a mass of 48g. How many moles of magnesium have you been given?

[Higher tier]

Answer 30

Number of moles = 48g ÷ 24 = 2 mol

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 31

Question:

You are given 120g of calcium. How many moles of calcium have you been given?

[Higher tier]

Answer 31

Number of moles = 120g ÷ 40 = 3 mol

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 32

Question:

A sample of rock contains 252g of iron. Calculate the number of moles of iron in the sample.

[Higher tier]

Answer 32

Number of moles = 252g ÷ 56 = 4.5 mol

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 33

Question:

You are given a sample of sulphur with a mass of 4064g. Calculate the number of moles of sulphur in the sample.

[Higher tier]

Answer 33

Number of moles = 4064g ÷ 32 = 127 mol

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 34

What is the equation for calculating the number of moles of a compound?

[Higher tier]

Answer 34

Number of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 35

Question:

You are given a sample of calcium carbonate (CaCO₃) with a mass of 300g. Calculate the number of moles of calcium carbonate in the sample.

[Higher tier]

Answer 35

Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100
Number of moles = 300g ÷ 100 = 3 mol

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 36

Question:

You are given 380g of magnesium chloride (MgCl₂). How many moles of magnesium chloride have you been given?

[Higher tier]

Answer 36

Mr of MgCl₂ = 24 + (35.5 × 2) = 95
Number of moles = 380g ÷ 95 = 4 mol

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 37

Question:

You are given a sample of lithium sulfate (Li₂SO₄) with a mass of 990g. Calculate the number of moles of lithium sulfate in the sample.

[Higher tier]

Answer 37

Mr of Li₂SO₄ = (7 × 2) + 32 + (16 × 4) = 110
Number of moles = 990g ÷ 110 = 9 mol

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 38

Question:

You are given 64.5g of beryllium hydroxide (Be(OH)₂). Calculate the number of moles of beryllium hydroxide that you have been given.

[Higher tier]

Answer 38

Mr of Be(OH)₂ = 9 + (16 × 2) + (1 × 2) = 43
Number of moles = 64.5g ÷ 43 = 1.5 mol

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 39

Rearrange the equation for calculating the number of moles of a compound and an element to make mass the subject.

[Higher tier]

Answer 39

• Mass (g) = number of moles (mol) × Relative formula mass or Mr
• Mass (g) = number of moles (mol) × Relative atomic mass or Ar

Question 40

Question:

Calcuate the mass of four moles of sodium chloride (NaCl).

[Higher tier]

Answer 40

Mr of NaCl = 23 + 35.5 = 58.5
Mass = 4 mol × 58.5 = 234g

Mass (g) = number of moles (mol) × Relative formula mass or Mr

Question 41

Question:

Calculate the mass of three moles of potassium oxide (K₂O).

[Higher tier]

Answer 41

Mr of K₂O = (39 × 2) + 16 = 94
Mass = 3 mol × 94 = 282g

Mass (g) = number of moles (mol) × Relative formula mass or Mr

Question 42

Question:

Calculate the mass of 0.1 moles of caesium nitrate (CsNO₃).

[Higher tier]

Answer 42

Mr of CsNO₃ = 133 + 14 + (16 × 3) = 195
Mass = 0.1 mol × 195 = 19.5g

Mass (g) = number of moles (mol) × Relative formula mass or Mr

Question 43

Question:

Calculate the mass of 5 moles of coper sulfate (CuSO₄).

[Higher tier]

Answer 43

Mr of CuSO₄ = 63.5 + 32 + (16 × 4) = 159.5
Mass = 5 mol × 159.5 = 797.5g

Mass (g) = number of moles (mol) × Relative formula mass or Mr

Question 44

Question:

2g of hydrogen reacts with 71g of chlorine to make 73g of hydrogen chloride. Using this information, balance the equation.

H₂ + Cl₂ ⭢ HCl

[Higher tier]

Answer 44

Moles of hydrogen = 2g ÷ 2 = 1 mol
Moles of chlorine = 71g ÷ 71 = 1 mol
Moles of hydrogen chloride = 73g ÷ 36.5 = 2 mol
| 1 ÷ 1 = 1 | 1 ÷ 1 = 1 | 2 ÷ 1 = 2 |

H₂ + Cl₂ ⭢ 2HCl

Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient

Question 45

Question:

54g of aluminium reacts with 216g of iron (II) oxide, forming 102g of aluminium oxide and 168g of iron. Using this information, balance the equation.

Al + FeO ⭢ Al₂O₃ + Fe

[Higher tier]

Answer 45

Moles of aluminium = 54g ÷ 27 = 2 mol
Moles of iron (II) oxide = 216g ÷ (56 + 16) = 3 mol
Moles of aluminium oxide = 102g ÷ ((27 × 2) + (16 × 3)) = 1 mol
Moles of iron = 168g ÷ 56 = 3 mol
| 2 ÷ 1 = 2 | 3 ÷ 1 = 3 | 1 ÷ 1 = 1 | 3 ÷ 1 = 3 |

2Al + 3FeO ⭢ Al₂O₃ + 3Fe

Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient

Question 46

Question:

1248g of barium chloride reacts with 684g of aluminium sulfate, forming 1398g of barium sulfate and 534g of aluminium chloride. Using this information, balance the equation.

BaCl₂ + Al₂(SO₄)₃ ⭢ BaSO₄ + AlCl₃

[Higher tier]

Answer 46

Moles of barium chloride = 1248g ÷ (137 + (35.5 × 2)) = 6 mol
Moles of aluminium sulfate = 684g ÷ ((27 × 2) + (32 × 3) + 3(16 × 4)) = 2 mol
Moles of barium sulfate = 1398g ÷ (137 + 32 + (16 × 4)) = 6 mol
Moles of aluminium chloride = 534g ÷ (27 + (35.5 × 3)) = 4 mol
| 6 ÷ 2 = 3 | 2 ÷ 2 = 1 | 6 ÷ 2 = 3 | 4 ÷ 2 = 2 |

3BaCl₂ + Al₂(SO₄)₃ ⭢ 3BaSO₄ + 2AlCl₃

Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient

Question 47

Question:

24g of magnesium reacts with 16g of oxygen, forming 40g of magnesium oxide. Using this information, balance the equation.

Mg + O₂ ⭢ MgO

[Higher tier]

Answer 47

Moles of magnesium = 24g ÷ 24 = 1 mol
Moles of oxygen = 16g ÷ (16 × 2) = 0.5 mol
Moles of magnesium oxide = 40g ÷ (24 + 16) = 1 mol
| 1 ÷ 0.5 = 2 | 0.5 ÷ 0.5 = 1 | 1 ÷ 0.5 = 2 |

2Mg + O₂ ⭢ 2MgO

Number of moles (mol) = Mass (g) ÷ Ar or Mr
Smallest coefficients = coefficients ÷ smallest coefficient

Question 48

What is Avogadro’s constant?

[Higher tier]

Answer 48

The number of atoms, molecules or ions in a mole of a given substance.

Question 49

What is the value of Avogadro’s constant?

[Higher tier]

Answer 49

6.02 × 10²³

Question 50

Question:

Calculate the number of moles of atoms in one mole of water molecules (H₂O).

Answer 50

1 molecule of H₂O = 3 atoms
∴ 1 mole of H₂O = 3 moles of atoms

Question 51

Question:

Calculate the number of moles of atoms in one mole of methane (CH₄).

[Higher tier]

Answer 51

1 molecule of CH₄ = 5 atoms
∴ 1 mole of CH₄ = 5 moles of atoms

Question 52

Question:

Calculate the number of moles of atoms in one mole of calcium hydroxide (Ca(OH)₂).

[Higher tier]

Answer 52

1 molecule of Ca(OH)₂ = 5 atoms
∴ 1 mole of Ca(OH)₂ = 5 moles of atoms

Question 53

Question:

Calculate the number of atoms in one mole of hydrogen chloride (HCl).

[Higher tier]

Answer 53

1 molecule of HCl = 2 atoms
1 mole of HCl = 6.02 × 10²³
Number of atoms in one mole 2 × 6.02 × 10²³ = 1.204 × 10²⁴

Question 54

Question:

Calculate the number of atoms in one mole of sodium oxide (Na₂O).

[Higher tier]

Answer 54

Number of atoms in 1 molecule of Na₂O = 3
1 mole = 6.02 × 10²³
Number of atoms in one mole = 3 × 6.02 × 10²³ = 1.806 × 10²⁴

Question 55

Question:

Calculate the number of atoms in 48g of magnesium (Mg).

[Higher tier]

Answer 55

Moles (mol) = 48g ÷ 24 = 2 mol
1 atom in one element of magnesium.
Number of atoms = 2 × 6.02 × 10²³

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 56

Question:

Calculate the number of atoms in 28g of lithium (Li).

[Higher tier]

Answer 56

Moles (mol) = 28g ÷ 7 = 4 mol
1 atom in one element of lithium.
Number of atoms = 4 mol × 6.02 × 10²³ = 2.408 × 10²⁴

# of moles (mol) = Mass (g) ÷ Relative atomic mass or Ar

Question 57

Question:

Calculate the number of atoms in 56g of calcium oxide (CaO).

[Higher tier]

Answer 57

Mole (mol) = 56g ÷ (40 + 16) = 1 mol
2 atoms in one molecule of calcium oxide.
Number of atoms = 2 × 6.02 × 10²³ = 1.204 × 10²⁴

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 58

Question:

Calculate the number of atoms in 54g of water (H₂O).

[Higher tier]

Answer 58

Moles = 54g ÷ ((1 × 2) + 16) = 3 mol
3 atoms in one molecule of water.
3 × 3 = 9
Number of atoms = 9 × 6.02 × 10²³ = 5.418 × 10²⁴

# of moles (mol) = Mass (g) ÷ Relative formula mass or Mr

Question 59

What are the steps to calculating the masses of products or reactants using reacting masses in a chemical equation?

[Higher tier]

Answer 59

1. Identify the known and unknown substances from the question.
2. Calculate Mr or Ar of the known substance using your periodic table.
3. Calculate moles of known substance using Mr (or Ar) and the given mass.
4. Workout the moles of the unknown substance using the ratio, which are the coefficients.
5. Calculate the Mr or Ar of the unknown substance using your periodic table.
6. Calculate the mass of the unknown substance using the moles and Mr (or Ar).

Question 60

Question:

Calculate the mass of magnesium chloride that could be produced from 72g of magnesium. Assume that the chlorine is unlimited.

Mg + Cl₂ ⭢ MgCl₂

[Higher tier]

Answer 60

1. Identify known & unknown substance.
   Known: Mg | Unknown: MgCl₂
2. Ar or Mr of known substance.
   Ar of Mg = 24
3. Moles of known substance.
   Moles (mol) = 72g ÷ 24 = 3 mol
4. Moles of unknown substance from ratio.
   Ratio = 1:1 | Moles (mol) = 3 moles
5. Mr or Ar of unknown substance.
   Mr of MgCl₂ = 95
6. Mass of unknown substance.
   Mass (g) = 3 mol × 95 = 285g

Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 61

Question:

Calculate the mass of calcium sulfate that could be produced from 80g of calcium. Assume that the sulfuric acid is unlimited.

Ca + H₂SO₄ ⭢ CaSO₄ + H₂

[Higher tier]

Answer 61

1. Identify known & unknown substance.
   Known: Ca | Unknown: CaSO₄
2. Ar or Mr of known substance.
   Ar of Ca = 40
3. Moles of known substance.
   Moles (mol) = 80g ÷ 40 = 2 mol
4. Moles of unknown substance from ratio.
   Ratio = 1:1 | Moles (mol) = 2 moles
5. Mr or Ar of unknown substance.
   Mr of CaSO₄ = 136
6. Mass of unknown substance
   Mass (g) = 2 mol × 136 = 272g

Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 62

Question:

Calculate the mass of calcium carbonate that we would need to produce 224g of calcium oxide.

CaCO₃ ⭢ CaO + CO₂

[Higher tier]

Answer 62

1. Identify known & unknown substance.
   Known: CaO | Unknown: CaCO₃
2. Mr or Ar of known substance.
   Mr of CaO = 56
3. Moles of known substance
   Moles (mol) = 224g ÷ 56 = 4 mol
4. Moles of unknown substance from ratio.
   Ratio: 1:1 | Moles (mol) = 4 moles
5. Mr or Ar of unknown substance.
   Mr of CaCO₃ = 100
6. Mass of unknown substance.
   Mass (g) = 4 mol × 100 = 400g

Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 63

Question:

Calculate the mass of magnesium chloride that could be produced from 146g of hydrochloric acid. Assume that the magnesium hydroxide is unlimited.

Mg(OH)₂ + 2HCl ⭢ MgCl₂ + H₂O

[Higher tier]

Answer 63

1. Identify known and unknown substance.
   Known: HCl | Unknown: MgCl₂
2. Mr or Ar of known substance.
   Mr of HCl = 36.5
3. Moles of known substance
   Moles (mol) = 146g ÷ 36.5 = 4 mol
4. Moles of unknown substance from ratio.
   Ratio = 2:1 | Moles (mol) = 4 ÷ (2 ÷ 1) = 2 moles
5. Mr or Ar of unknown substance.
   Mr of MgCl₂ = 95
6. Mass of unknown substance.
   Mass (g) = 2 mol × 95 = 190g

Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 64

Question:

Calculate the mass of sodium sulfate that could be produced from 240g of sodium hydroxide. Assume that the sulfuric acid is unlimited.

2NaOH + H₂SO₄ ⭢ Na₂SO₄ + 2H₂O

[Higher tier]

Answer 64

1. Identify known and unknown substance.
   Known: NaOH | Unknown: Na₂SO₄
2. Mr or Ar of known substance.
   Mr of NaOH = 40
3. Moles of known substance
   Moles (mol) = 240g ÷ 40 = 6 mol
4. Moles of unknown substance from ratio.
   Ratio = 2:1 | Moles (mol) = 6 ÷ (2 ÷ 1) = 3 moles
5. Mr or Ar of unknown substance.
   Mr of Na₂SO₄ = 142
6. Mass of unknown substance.
   Mass (g) = 3 mol × 142 = 426g

Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 65

Question:

Calculate the mass of hydrogen peroxide that could produce 64g of oxygen.

2H₂O₂ ⭢ 2H₂O + O₂

[Higher tier]

Answer 65

1. Identify known and unknown substance.
   Known: O₂ | Unknown: H₂O₂
2. Mr or Ar of known substance.
   Ar of O₂ = 16 × 2 = 32
3. Moles of known substance
   Moles (mol) = 64g ÷ 32 = 2 mol
4. Moles of unknown substance from ratio.
   Ratio = 1:2 | Moles (mol) = 2 × 2 = 4 moles
5. Mr or Ar of unknown substance.
   Mr of H₂O₂ = 34
6. Mass of unknown substance.
   Mass (g) = 4 mol × 34 = 136g

Moles (mol) = mass (g) ÷ relative atomic mass or Ar
Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 66

Question:

Nitrogen and hydrogen form ammonia shown by the following equation:

N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia.

[Higher tier]

Answer 66

1. Identify known and unknown substance.
   Known: NH₃ | Unknown: N₂
2. Mr of known substance.
   Mr of NH₃ = 17
3. Convert mass into grams
   1 tonne = 1000kg = 1 × 10⁶g | Mass = 6.8 tonnes ⤚ ^{(×1,000,000)} ⭢ 6,800,000g
4. Moles of known substance.
   Moles (mol) = (6.8 × 10⁶)g ÷ 17 = 400,000 mol
5. Moles of unknown from ratio.
   Ratio = 2:1 | Moles = 400,000 ÷ (2 ÷ 1) = 200,000 moles
6. Mr of unknown substance
   Mr of N₂ = 28
7. Mass of unknown substance.
   Mass (g) = 200,000 mol × 28 = 5,600,000g (5.6 tonnes)

Moles (mol) = mass (g) ÷ relative formula mass or Mr

Question 67

What is a limited reactant (or reagent)?

[Higher tier]

Answer 67

The reactant (in a chemical equation) that is completely used up, limiting the amount of products produced.

Question 68

What is the excess reactant (or reagent)?

[Higher tier]

Answer 68

The reactant or reagent that is left after a chemical reaction is complete.

Question 69

What are the steps to calculating the masses of products or reactants using limited reactants in a chemical equation?

[Higher tier]

Answer 69

1. Identify and ignore, for the rest of your calculations, the excess reactant.
2. Identify the known and unknown substance from the limited reactant and what you’re trying to find.
3. Find the moles of product from moles of limited reactant (they are the same).
4. Find the mass of the product using moles and Mr or Ar.

Question 70

Question:

How many moles of zinc iodide would be produced if we used 0.5 moles of zinc and 1 mole of iodine? Calculate the mass of product.

Zn + I₂ ⭢ ZnI₂

[Higher tier]

Answer 70

1. Identify and ignore the excess reactant.
   Given moles: 1 > 0.5 | Excess reactant = I₂
2. Identify known and unknown substance.
   Known: Zn | Unkown: ZnI₂
3. Find moles of product from limited reactant.
   Moles of Zn = moles of ZnI₂ = 0.5 mol
4. Mass of product.
   Mass (g) = 0.5 mol × (65 + (127 × 2)) = 159.5g

Mass (g) = number of moles (mol) × relative formula mass or Mr

Question 71

Question:

How many moles of sodium chloride will be produced if we used 1 mole of sodium hydroxide and 0.25 moles of hydrochloric acid? Calculate the mass of sodium chloride produced.

NaOH + HCl ⭢ NaCl + H₂O

[Higher tier]

Answer 71

1. Identify and ignore the excess reactant.
   Given moles: 1 > 0.25 | Excess = NaOH
2. Identify known and unknown substance.
   Known: NaOH | Unknown: NaCl
3. Moles of product from limited reactant.
   Moles of NaOH = moles of NaCl | Moles = 0.25 mol
4. Mass of product.
   Mass (g) = 0.25 mol × (23 + 35.5) = 14.625g

Mass (g) = number of moles (mol) × relative formula mass or Mr

Question 72

Question:

How many moles of copper will be produced if we used 0.5 moles of copper sulphate and 1 mole of magnesium? Calculate the mass of copper produced.

CuSO₄ + Mg ⭢ MgSO₄ + Cu

[Higher tier]

Answer 72

1. Identify and ignore the excess reactant.
   Given moles: 0.5 > 1 | Excess = Mg
2. Identify known and unknown substance.
   Known: CuSO₄ | Unknown: Cu
3. Moles of product from limited reactant.
   Moles of CuSO₄ = moles of Cu | Moles = 0.5 mol
4. Mass of product.
   Mass (g) = 0.5 mol × 63.5 = 31.75g

Mass (g) = number of moles (mol) × relative atomic mass or Ar

Question 73

What is meant by concentration; what is the unit?

[Foundation & Higher tier]

Answer 73

Concentration tells us the mass of a solute in a given volume of solution; g/dm³.

Question 74

What is a solute?

Answer 74

A chemical that is dissolved in a solvent.

Question 75

What is the equation for calculating the concentration of a solution?

(not in specification but worth learning)

Answer 75

Concentration (g/dm³) = mass (g) ÷ volume (dm³)

Question 76

Question:

200g of a chemical is dissolved in water to a final volume of 1 dm³. Calculate the concentration of the solution.

Answer 76

Concentration = 200g ÷ 1 dm³ = 200g/dm³

Concentration (g/dm³) = mass (g) ÷ volume (dm³)

Question 77

Question:

150g of a chemical is dissolved in water to a final volume of 0.5 dm³. Calculate the concentration of the solution.

Answer 77

Concentration (g/dm³) = 150g ÷ 0.5 dm³ = 300 g/dm³

Concentration (g/dm³) = mass (g) ÷ volume (dm³)

Question 78

Quetsion:

Calculate the mass of a chemical needed to dissolve in a final volume of 0.4 dm³ to give a final concentration of 600 g/dm³.

Answer 78

Mass (g) = 600 g/dm³ × 0.4 dm³ = 240g

Mass (g) =| Concentration (g/dm³) × volume (dm³)

Question 79

Question:

Calculate the final volume of a solution containing 200g of a chemical with a concentration of 800 g/dm³.

Answer 79

Volume (dm³) = 200g ÷ 800g/dm³ = 0.25 dm³

Volume (dm³) = mass (g) ÷ concentration (g/dm³)

Question 80

What 2 factors affect the concentration?

[Higher tier]

Answer 80

• Mass of the solute: increase in mass = increase in concentration (if volume of solution kept the same).
• Volume of the solution: increase in volume = decrease in concentration (if mass of solute kept the same).

Question 81

What is yield?

Triple only

Answer 81

The mass of product that a chemical reaction produces.

Question 82

What is the problem with calculating the yield, in practice?

Triple only

Answer 82

It is not always possible to achieve 100% yield in a chemical reaction.

Question 83

What are the reasons why you cannot always achieve 100% yield in a chemical reaction? (3)

Triple only

Answer 83

1. Some of the product may be lost when it is separated from the reaction mixture.
2. Some of the reactants may react in different ways to the expected reaction so we do not get the product we expect.
3. Reversible reactions may not go to completion.

Question 84

What is the equation for calculating the percentage yield?

Triple only

Answer 84

Percentage yield = 100 (mass of product actually made ÷ maximum theoretical mass of product)

Question 85

What are the steps to calculating the percentage yield?

Triple only

Answer 85

1. Find out the theoretical mass by calculating the mass of the product provided in the question, using reacting masses.
2. Calculate the percentage yield using your calculated maximum theoretical mass and the mass of the product given in the question.

Question 86

Question:

A scientist reacted 48g of magnesium and produced 150g of magnesium sulphate. Calculate the percentage yield.

Mg + H₂SO₄ ⭢ MgSO₄ + H₂

Triple only

Answer 86

Step 1:

1. Identify known and unknown substances
   Known: Mg | Unknown: MgSO₄
2. Mr of known and unknown substance.
   Ar of Mg = 24 | Mr of MgSO₄ = 120
3. Moles of known substance.
   Moles (mol) = 48g ÷ 24 = 2 mol
4. Mole of unknown from ratio.
   Ratio = 1:1 | Moles (mol) = 2 mol
5. Mass of unknown.
   [Theoretical] mass (g) = 2 mol × 120 = 240 g

Step 2:
Percentage yield = 150g ÷ 240g = 0.625 × 100 = 62.5%

• Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
• % yield = 100 (actual mass ÷ theoretical mass)

Question 87

Question:

The theoretical yield of beryllium chloride was 10.7g. If the reaction actually yields 4.5g, what was the percent yield?

Be + 2HCl ⭢ BeCl₂ + H₂

Triple only

Answer 87

Percentage yield = (4.5g ÷ 10.7g) × 100 = 0.42 × 100 = 42%

% yield = 100 (actual mass ÷ theoretical mass)

Question 88

Question:

2a) A scientist began this reaction with 20 grams of lithium hydroxide and an unlimited amount of KCl.
What is the theoretical yield of lithium chloride (grams)?

LiOH + KCl ⭢ LiCl + KOH

2b) The reaction actually produced 6 grams of lithium chloride. What is the percent yield?

Triple only

Answer 88

Part A:

1. Identify known and unknown substances.
   1Known: LiOH | Unknown: LiCl
2. Mr of known and unknown substance.
   Mr of LiOH = 24 | Mr of LiCl = 42.5
3. Moles of known substance.
   Moles (mol) = 20g ÷ 24 = 0.8 mol
4. Mole of unknown from ratio.
   Ratio = 1:1 | Moles (mol) = 0.8 mol
5. Mass of unknown.
   [Theoretical] mass (g) = 0.8 mol × 42.5 = 34 g

Part B:
Percentage yield = 6g ÷ 34g = 0.176 × 100 = 17.6%

• Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
• % yield = 100 (actual mass ÷ theoretical mass)

Question 89

Question:

Nitrogen and hydrogen at high temperatures are converted to ammonia using the following reaction known as the Haber Process.

3H₂ + N₂ ⭢ 2NH₃

When 400 grams of H₂ are added to an excess amount of N₂, 104 grams of NH₃ are formed. Calculate the percent yield.

Triple only

Answer 89

Step 1:

1. Identify known and unknown substances.
   Known: H₂ | Unknown: NH₃
2. Mr of known and unknown substance.
   Mr of H₂ = 2 | Mr of NH₃ = 17
3. Moles of known substance.
   Moles (mol) = 400g ÷ 2 = 200 mol
4. Mole of unknown from ratio.
   Ratio = 3:2 | Moles = 200 ÷ (3 ÷ 2) = 133.3 mol
5. Mass of unknown.
   [Theoretical] mass (g) = 133.3 mol × 17 = 2266g

Step 2:
Percentage yield = 104g ÷ 2266g = 0.046 × 100 = 4.6%

• Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
• % yield = 100 (actual mass ÷ theoretical mass)

Question 90

Question:

If the typical yield is 86.78%, how much SO₂ should be expected if 4897 grams of ZnS are used? (2803 of SO₂)

ZnS + O₂ ⭢ ZnO + SO₂

Triple only

Answer 90

Step 1:

1. Identify known and unknown substances.
   Known: ZnS | Unknown: SO₂
2. Mr of known and unknown substance.
   Mr of ZnS = 97 | Mr of SO₂ = 64
3. Moles of known substance.
   Moles (mol) = 4897g ÷ 97 = 50.5 mol
4. Mole of unknown from ratio.
   Ratio = 1:1 | Moles = 50.5 mol
5. Mass of unknown.
   [Theoretical] mass (g) = 50.5 mol × 64 = 3232g

Step 2:
Actual mass = 86.78% × 3232g = 2804.7g

• Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
• Actual mass = % yield × theoretical mass

Question 91

Question:

A scientist reacted 130g of zinc with unlimited copper sulfate. They produced 25.4g of copper. Calculate the percentage yield.

Zn + CuSO₄ ⭢ Cu + ZnSO₄

Triple only

Answer 91

Step 1:

1. Identify the known and unknown.
   Known: Zn | Unknown: Cu
2. Ar of known and unknown.
   Zn = 65 | Cu = 63.5
3. Moles of known.
   Moles = 130g ÷ 65 = 2 mol
4. Moles of unknown from ratio.
   Ratio = 1:1 | Moles = 2 mol
5. Mass of unknown.
   [Theoretical] mass = 2 mol × 63.5 = 127g

Step 2:
Percentage yield = 25.4g ÷ 127g = 0.2 × 100 = 20%

• Number of moles (mol) = mass (g) ÷ relative atomic mass or Ar
• % yield = actual mass ÷ theoretical mass

Question 92

Question:

A scientist reacted 27g of beryllium with unlimited hydrochloric acid. They produced 144g of beryllium chloride. Calculate the percentage yield.

Be + 2HCl ⭢ BeCl₂ + H₂

Triple only

Answer 92

Step 1:

1. Identify known and unknown.
   Known: Be | Unknown: BeCl₂
2. Mr and Ar of known and unknown.
   Be = 9 | BeCl₂ = 80
3. Moles of known.
   Moles = 27g ÷ 9 = 3 mol
4. Find moles of unknown from ratio.
   Ratio = 1:1 | Moles = 3 mol
5. Mass of unknown.
   [Theoretical] mass = 3 mol × 80 = 240g

Step 2:
Percentage yield = 144g ÷ 240g = 0.6 × 100 = 60%

• Number of moles (mol) = mass (g) ÷ relative atomic mass or Ar
• % yield = actual mass ÷ theoretical mass

Question 93

What is atom economy

Triple only

Answer 93

A measure of the amount of starting materials that end up as useful products.

Question 94

Why is a high atom economy important in industry?

Triple only

Answer 94

• By minimising the production of unwanted products, money is saved.
• By minimising the production of unwanted products, sustainability increases as resources are not wasted.

Question 95

What is the equation for calculating atom economy?

Triple only

Answer 95

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 96

What is the greatest atom economy we can have? Why?

Triple only

Answer 96

100% because anything greater would mean we created atoms, which is not possible.

Question 97

Question:

Calculate the percentage atom economy for the production of silver iodide in this reaction.

The equation for the reaction is:

AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq)

Give your answer to 3 significant figures.

Relative formula masses:
AgNO₃ = 170 | NaI = 150 | AgI = 235 | NaNO₃ = 85

Triple only

Answer 97

Atom economy = 235 ÷ 320 = (0.7340) × 100 = 73.4%

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 98

Question:

This question is about the extraction of metals.

Tungsten (W) is a metal. Tungsten is produced from tungsten oxide by reaction with hydrogen.

The equation for the reaction is:
WO₃ + 3H₂ → W + 3H₂O

Calculate the percentage atom economy when tungsten is produced in this reaction.

Relative formula masses (Mr):
WO₃ = 232 | H₂ = 2

Triple only

Answer 98

Ar of W = 184
Atom economy = 184 ÷ (232 + (3 × 2)) = 0.773 × 100 = 77.3%

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 99

How do chemists try to reduce waste when unwanted side products have been produced?

Answer 99

They use the side products in other reactions.

Question 100

Question:

This question is about reversible reactions and equilibrium.
Hydrogen is used to produce ammonia in the Haber process.
The hydrogen is made in two stages.
Stage 1 is the reaction of methane and steam to produce carbon monoxide and hydrogen.

The equation for the reaction is:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)

Calculate the atom economy for the formation of hydrogen in stage 1.

Relative atomic masses (Ar): H = 1 | C = 12 | O = 16

Triple only

Answer 100

Atom economy = 3(1 × 2) ÷ 34 = 0.17647 × 100 = 17.7%

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 101

Question:

An equation for the reaction is:
NiO + C ⟶ Ni + CO

Calculate the percentage atom economy for the reaction to produce nickel.

Relative atomic masses (Ar): C = 12 | Ni = 59
Relative formula mass (Mr): NiO = 75
Give your answer to 3 significant figures.

Triple only

Answer 101

Atom economy = 59 ÷ (75 + 12) = 0.6781 × 100 = 67.8%

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 102

Question:

(d) Look at the equations for the two reactions:
Reaction 1: CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) + CO₂(g)
Reaction 2: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l)

Calculate the percentage atom economy for Reaction 2 [Copper chloride].
Reactive formula masses: CuO = 79.5 | HCl = 36.5 | CuCl₂ = 134.5 | H₂O = 18

(e) The atom economy for Reaction 1 is 68.45%.
Compare the atom economies of the two reactions for making copper chloride.
Give a reason for the difference.

Triple only

Answer 102

Part D:
Atom economy = 134.5 ÷ (79.5 + (2 × 36.5)) = 0.882 × 100 = 88.2%

Part E:
Reaction 1 has a lower atom economy than reaction 2 because more side products were produced.

Atom economy = (Mr (involving coefficients) of desired products ÷ sum of Mr (involving coefficients) of all reactants) × 100

Question 103

What is meant by concentration; what is the unit?

Triple only

Answer 103

Concentration tells us the number of moles of a solute in a given volume of solution; mol/dm³.

Question 104

What is the equation for calculating the concentration of a solution?

Triple only

Answer 104

Concentration (mol/dm³) = number of moles ÷ volume (dm³)

Question 105

Question:

A solution has a concentration of 0.5 mol/dm³. Calculate the number of moles in 0.2 dm³.

Triple only

Answer 105

Number of moles (mol) = 0.5 mol/dm³ × 0.2 dm³ = 0.01 moles

# of moles = concentration (mol/dm³) × volume (dm³)

Question 106

Question:

A solution has a concentration of 0.2 mol/dm³. Calculate the number of moles in 1.5 dm³.

Triple only

Answer 106

Number of moles = 0.2 mol/dm³ × 1.5 dm³ = 0.3 moles

# of moles = concentration (mol/dm³) × volume (dm³)

Question 107

Question:

A solution of calcium chloride has a concentration of 0.4 mol/dm³. Calculate the mass of calcium chloride in 2 dm³.

Mr: CaCl₂ = 111

Triple only

Answer 107

Step 1: Calculate the number of moles.
Number of moles = 0.4 mol/dm³ × 2 dm³ = 0.8 moles

Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.8 moles × 111 = 88.8g (1.d.p)

# of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass or Mr

Question 108

Question:

A solution of sodium nitrate has a concentration of 0.8 mol/dm³. Calculate the mass of sodium nitrate in 0.5 dm³.

Mr: NaNO₃ = 85

Triple only

Answer 108

Step 1: Calculate the number of moles.
Number of moles = 0.8 mol/dm³ × 0.5 dm³ = 0.4 moles

Step 2: Calculate the mass from moles and Mr.
Mass (g) = 0.4 moles × 85 = 34g

# of moles = concentration (mol/dm³) × volume (dm³)
Mass (g) = # of moles (mol) × Relative formula mass or Mr

Question 109

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“Using concentration of solution 2 (Triple)”