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Upcoming test Flashcards

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1
Q

Question:

Balance the equation:

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

A

CaO + 2HCl ⭢ CaCl₂ + H₂O

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2
Q

Question:

Balance the equation:

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

A

Fe₂O₃ + 3CO ⭢ 2Fe + 3CO₂

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3
Q

What is the Relative Atomic Mass (Ar)?

A

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).

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4
Q

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

A

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

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5
Q

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

A
  1. They have no units
  2. Do not take into account coefficients (big numbers) when calculating the Mr.
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6
Q

Question:

Calculate the Mr of CaSO₄

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

A

Mr = 40 + 32 + (16 × 4) = 136

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7
Q

Question:

Calculate the Mr of Mg(OH)₂

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

A

Mr = 24 + (16 × 2) + (1 × 2) = 58

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8
Q

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

[Higher tier]

A

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

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9
Q

Question:

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

[Higher tier]

A

Number of moles = 48g ÷ 24 = 2 mol

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

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10
Q

Question:

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

[Higher tier]

A

Number of moles = 120g ÷ 40 = 3 mol

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

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11
Q

Question:

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

[Higher tier]

A

Number of moles = 252g ÷ 56 = 4.5 mol

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

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12
Q

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]

A

Number of moles = 4064g ÷ 32 = 127 mol

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

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13
Q

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

[Higher tier]

A

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

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14
Q

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]

A

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

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15
Q

Question:

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

[Higher tier]

A

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

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16
Q

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]

A

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

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17
Q

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]

A

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

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18
Q

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

[Higher tier]

A
  • Mass (g) = number of moles (mol) × Relative formula mass or Mr
  • Mass (g) = number of moles (mol) × Relative atomic mass or Ar
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19
Q

Question:

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

[Higher tier]

A

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

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20
Q

Question:

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

[Higher tier]

A

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

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

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21
Q

Question:

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

[Higher tier]

A

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

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22
Q

Question:

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

[Higher tier]

A

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

23
Q

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]

A

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

24
Q

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]

A

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

25
Q

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]

A

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

26
Q

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]

A

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

27
Q

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

[Higher tier]

A
  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).
28
Q

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]

A
  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

29
Q

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]

A
  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

30
Q

Question:

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

CaCO₃ ⭢ CaO + CO₂

[Higher tier]

A
  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

31
Q

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]

A
  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

32
Q

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]

A
  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

  1. Mr or Ar of unknown substance.
    Mr of Na₂SO₄ = 142
  2. Mass of unknown substance.
    Mass (g) = 3 mol × 142 = 426g

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

33
Q

Question:

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

2H₂O₂ ⭢ 2H₂O + O₂

[Higher tier]

A
  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

34
Q

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]

A
  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

35
Q

What is meant by concentration; what is the unit?

[Foundation & Higher tier]

A

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

36
Q

What is a solute?

A

A chemical that is dissolved in a solvent.

37
Q

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

(not in specification but worth learning)

A

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

38
Q

Question:

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

A

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

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

39
Q

Question:

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

A

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

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

40
Q

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.

A

Mass (g) = 600 g/dm³ × 0.4 dm³ = 240g

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

41
Q

Question:

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

A

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

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

42
Q

What 2 factors affect the concentration?

[Higher tier]

A
  • 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).
43
Q

What is yield?

Triple only

A

The mass of product that a chemical reaction produces.

44
Q

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

Triple only

A

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

45
Q

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

Triple only

A
  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.
46
Q

Question:

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

Mg + H₂SO₄ ⭢ MgSO₄ + H₂

Triple only

A

1.1. Identify known and unknown substances
Known: Mg | Unknown: MgSO₄

1.2. _

47
Q

What is the equation for calculating the percentage yield?

Triple only

A

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

48
Q

What are the steps to calculating the percentage yield?

Triple only

A
  1. Find out the theoretical mass by calculating the mass of the product(s) or reactant(s), asked in the question (reaction mass Q.), using reacting masses.
  2. Calculate the percentage yield using your calculated maximum theoretical mass and the mass given in the question (percentage yield Q.).
49
Q

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

A

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

% yield = 100 (actual mass ÷ predicted mass)

50
Q

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

A

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.
    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 ÷ predicted mass)
51
Q

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

A

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.
    Mass (g) = 133.3 mol × 17 = 2266g

Part B:
Percentage yield = 104g ÷ 2266g = 0.046 × 100 = 4.6%

  • Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
  • % yield = 100 (actual mass ÷ predicted mass)
52
Q

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

A

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.
    Mass (g) = 50.5 mol × 64 = 3232g

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

  • Number of moles (mol) = mass (g) ÷ relative formula mass or Mr
  • Actual mass = (% yield × predicted mass)
53
Q

Question:

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

Mg + H₂SO₄ ⭢ MgSO₄ + H₂

Triple only

A

Step 1:
1.1. Identify known and unknown substances
Known: Mg | Unknown: MgSO₄

1.2. Mr of known and unknown substance.
Ar of Mg = 24 | Mr of MgSO₄ = 120

1.3. Moles of known substance.
Moles (mol) = 48g ÷ 24 = 2 mol

1.4. Mole of unknown from ratio.
Ratio = 1:1 | Moles (mol) = 2 mol

1.5. Mass of unknown.
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 ÷ predicted mass)

Chemistry (16 decks)

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