Topic 3: Quantitive Chemistry

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 atomic mass and relative formula mass.

Answer 3

RAM - average mass of atoms in an element taking into account masses and abundance of its isotopes, relative to ^12C.

RFM - sum of RAM’s of all atoms in the formula.

Question 4

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

Answer 4

Mass = Mr x Moles

Question 5

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

Answer 5

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

Question 6

Calculate the amount of carbon dioxide in moles in 0.32 g of carbon dioxide.
Relative atomic masses (A,):

carbon =12, oxygen = 16

Answer 6

Moles = Mass / Mr
0.32 / 44 = 0.007

Question 7

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 (A,): H = 1; N = 14

Answer 7

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 × 28 = 5600000 g = 5.6 tonnes.

Question 8

State what we mean by a limiting reactant in a chemical reaction

Answer 8

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 9

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 9

Step 1: Write the balanced equation 2H2O2 → 2H2O + O2
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

Question 10

Write down the two formulae that link concentration, mass and volume together.

Answer 10

Concentration (g per dm3) = Mass (g)/Volume (dm3)

Concentration (mol per dm3)= nr of moles/volume (dm3)