C3 Quantitative Chemistry (Part 1)

Question 1

What is the law of conservation of mass?

Answer 1

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.

Question 2

Write a balanced equation of magnesium reacting with hydrochloric acid.

Answer 2

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

Question 3

Define relative formula mass.

Answer 3

RFM - sum of relative atomic masses of all atoms in the formula.

Question 4

How can you increase the precision of experiment results?

Answer 4

Measure to more decimal places or use a more sensitive balance / apparatus

Question 5

What is Avogadro’s constant?

Answer 5

The number of atoms, molecules or ions in a mole of a given substance. The value of the constant is 6.02 x 1023.

Question 6

What is the formula that links mass, molecular mass and moles together

Answer 6

Mass = Mr x Moles

Question 7

What is the mass of: 20 moles of calcium carbonate, CaCO3

Answer 7

Mass = Mr x Moles
Mr = 100
100 x 20 = 2000 g

Question 8

Calculate the amount of carbon dioxide
in moles in 0.32 g of carbon dioxide.

Relative atomic masses (Ar):
carbon = 12
oxygen = 16

Answer 8

CO2 = 12 + 16x2 = 44

Moles = Mass / Mr
0.32 / 44 = 0.007

Question 9

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

Answer 9

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.

Question 10

Why do we have a reactant in excess?

What is a limiting reactant?

Answer 10

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.

Question 11

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?

Answer 11

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