Reacting Masses

Extended tier only

Chemical / symbol equations can be used to calculate:
- The moles of reactants and products
- The mass of reactants and products
To do this:
- Information from the question is used to find the amount in moles of the substances being considered
- Then, the ratio between the substances is identified using the balanced chemical equation
- Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses

Magnesium undergoes combustion to produce magnesium oxide.

The overall reaction that is taking place is shown in the equation below.

2Mg (s) + O₂ (g) ⟶ 2 MgO (s)

Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction:

A_r(O) = 16 A_r(Mg) = 24

Answer:

Calculate the moles of magnesium
- Moles = $\frac{mass}{M_{r}}$ = $\frac{6}{24}$ = 0.25
Use the molar ratio from the balanced symbol equation
- 2 moles of magnesium produce 2 moles of magnesium oxide
- The ratio is 1 : 1
- Therefore, 0.25 moles of magnesium oxide is produced
Calculate the mass of magnesium oxide
- Mass = moles x M_r = 0.25 moles x (24 + 16) = 10 g

In theory, aluminium could decompose as shown in the equation below.

2Al₂O₃ ⟶ 4Al + 3O₂

Calculate the maximum possible mass of aluminium, in tonnes, that can be produced from 51 tonnes of aluminium oxide.

A_r(O) = 16 A_r(Al) = 27

Answer:

Calculate the moles of aluminium oxide
- Mass = 51 tonnes x 10⁶ = 51 000 000 g
- M_r of aluminium oxide = (2 x 27) + (3 x 16) = 102
- Moles = $\frac{mass}{M_{r}}$ = $\frac{51 000 000}{102}$ = 500 000
Use the molar ratio from the balanced symbol equation
- 2 moles of aluminium oxide produces 4 moles of aluminium
- The ratio is 1 : 2
- Therefore, 2 x 500 000 = 1 000 000 moles of aluminium is produced
Calculate the mass of aluminium
- Mass = moles x M_r = 1 000 000 moles x 27 = 27 000 000 g
- Mass in tonnes = $\frac{27 000 000}{10^{6}}$ = 27 tonnes

Remember: The molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.

A chemical reaction stops when one of the reactants is used up
The reactant that is used up first is the limiting reactant, as it limits the duration and hence the amount of product that a reaction can produce
- The one that is remaining is the excess reactant
- The limiting reagent is the reactant which is not present in excess in a reaction
The amount of product is therefore directly proportional to the amount of the limiting reactant added at the beginning of a reaction

In order to determine which reactant is the limiting reagent in a reaction, we have to consider the amounts of each reactant used and the molar ratio of the balanced chemical equation
When performing reacting mass calculations, the limiting reagent is always the number that should be used, as it indicates the maximum possible amount of product that can form
- Once all of a limiting reagent has been used up, the reaction cannot continue
The steps are:
1. Convert the mass of each reactant into moles by dividing by the molar masses
2. Write the balanced equation and determine the molar ratio
3. Look at the equation and compare the moles

9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na₂S.

Which reactant is in excess and which is the limiting reactant?

Relative atomic masses (A_r): Na = 23; S = 32

Answer:

Calculate the moles of each reactant
- Moles = $\frac{mass}{M_{r}}$
- Moles Na = $\frac{9.2}{23}$ = 0.40
- Moles S = $\frac{8.0}{32}$ = 0.25
Write the balanced equation and determine the molar ratio
- 2Na + S → Na₂S
- So, the molar ratio of Na : S is 2 : 1
Compare the moles
- To react completely 0.40 moles of Na requires 0.20 moles of S
- Since there are 0.25 moles of sulfur:
  - S is in excess
  - Therefore, Na is the limiting reactant

Reacting Masses (CIE IGCSE Chemistry: Co-ordinated Sciences (Double Award))