Reacting Masses (CIE IGCSE Chemistry)
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Reacting Masses
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- 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
Example 1
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
Example 2
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
Exam Tip
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
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- 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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