Reacting Masses (Cambridge (CIE) O Level Chemistry): Revision Note
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Reacting Masses
Chemical equations can be used to calculate the moles or masses of reactants and products
To do this, information given in 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
Worked Example
Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction:
2Mg (s) + O2 (g) ⟶ 2 MgO (s)
Relative formula masses (Mr): Mg = 24; MgO = 40
Worked Example
Calculate the mass of aluminium, in tonnes, that can be produced from 51 tonnes of aluminium oxide. The equation for the reaction is:
2Al2O3 ⟶ 4Al + 3O2
Relative formula masses (Mr): Al = 27; Al2O3 = 102
Examiner Tips and Tricks
Remember molar ratio of a balanced equation gives you the ratio of the amounts of each substance in the reaction.
Limiting Reactants
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 amount of product is therefore directly proportional to the amount of the limiting reactant added at the beginning of a reaction
The limiting reactant is the reactant which is not present in excess in a reaction
In order to determine which reactant is the limiting reactant in a reaction, we have to consider the ratios of each reactant in the balanced equation
When performing reacting mass calculations, the limiting reactant is always the number that should be used as it indicates the maximum possible amount of product
The steps are:
Write the balanced equation for the reaction
Calculate the moles of each reactant
Compare the moles & deduce the limiting reactant
Worked Example
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 (Ar): Na = 23; S = 32
Answer
Step 1: Write the balanced equation and determine the molar ratio
2Na + S → Na2S so the molar ratio of Na : S is 2 : 1
Step 2: Calculate the moles of each reactant
Moles = Mass ÷ Molar Mass
Moles Na = 9.2 ÷ 23 = 0.40
Moles S = 8.0 ÷ 32 = 0.25
Step 3: Compare the moles
To react completely 0.40 moles of Na requires 0.20 moles of S and since there are 0.25 moles of S, then S is in excess
Na is therefore the limiting reactant
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