Topic 3: Quantitative Chemistry Flashcards
What is the law of conservation of mass?
The law of conservation of mass states that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.
Write a balanced equation of magnesium reacting with hydrochloric acid.
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
Define relative atomic mass
RAM - average mass of atoms in an element taking into account masses and abundance of its isotopes, relative to 12C.
Define relative formula mass.
RFM - sum of RAM’s of all atoms in the formula.
What is the relative formula mass of:
A) CaF2
B) C6H12O6
CaF2 - (Ar values: Ca = 40, F = 19)
40 + 19 + 19 = 78
—
C2H12O6 - (Ar values: C = 12, H = 1, O = 16)
(12 x 6) + (1 x 12) + (16 x 6) = 180
The following reaction occurs in a test tube under a Bunsen Burner:
4 MgO(s) + CH4(g) → 4 Mg(s) + 2 H2O(g) + CO2(g)
The carbon dioxide and water escape from the test tube. Use the equation to explain why.
They are both gases
What is Avogadro’s constant?
The number of atoms, molecules or ions in a mole of a given substance. The value of the constant is 6.02 x 10^23.
What is the formula that links mass, molecular mass and moles together
Mass = Mr x Moles
What is the mass of: 20 moles of calcium carbonate, CaCO3
Mass = Mr x Moles
Mr = 100
100 x 20 = 2000 g
Calculate the amount of carbon dioxide in moles in 0.32 g of carbon dioxide.
Relative atomic masses (Ar): carbon = 12, oxygen = 16
Moles = Mass / Mr
0.32 / 44 = 0.007
Nitrogen and hydrogen form ammonia shown by the following equation:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia.
Relative atomic masses (Ar): H = 1; N = 14
- Step 1 - Work out the number of number of moles of ammonia (Mr of ammonia = 17)
6800000 / 17 = 400000 moles of ammonia
- Step 2 - Use the balanced equation and number of moles of ammonia to work out the number of moles of nitrogen
The ratio of nitrogen to ammonia is 1:2
Therefore the number of moles of nitrogen is 400000/2 = 200000
- Step 3 - Work out the mass of nitrogen (Mr of N2 is 28)
200000 x 28 = 5600000 g = 5.6 tonnes.
State what we mean by a limiting reactant in a chemical reaction
In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant because it limits the amount of products.
Hydrogen peroxide decomposes in water to form water and oxygen. How many grams of oxygen gas will be given off from 40.8 g of hydrogen peroxide?
- Step 1: Write the balanced equation 2 H2O2(l) → 2 H2O + O2(g)
Mr of H2O2 = 34 - Step 2: Number of moles in 40.8 g : 40.8/34 = 1.2 moles
Ratio in the balanced equation of H2O2 : O2 = 2:1
- Step 3 :Therefore number of moles of O2 = 0.6 moles
- Step 4: Mass of oxygen = 0.6 x 32 (Mr of O2) = 19.2
Write down the two formulae that link concentration, mass and volume together.
- Concentration (g per dm3) = Mass (g)/Volume (dm^3)
- Concentration (mol per dm^3) = nr of moles/volume (dm^3)
31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.
HNO3 + KOH → KNO3 + H2O
Calculate the concentration of the potassium hydroxide solution in moldm^-3
- Step 1: Calculate the moles of HNO3 used = Concentration x volume
2 x 0.025 dm3 (25/1000 to convert the units) = 0.05 moles
- Step 2 : Calculate the moles of KOH
Ratio is 1:1 therefore number of moles of KOH = 0.05 - Step 3 : Calculate the concentration of KOH Volume = Moles/concentration; 0.05 / 0.031 = 1.61