Chemical Formulae, Equations and Calculations Part 1
3 marks · NormalThe equation for a reaction is:
A student reacts 27.0 g of aluminium (Al) with 85.0 g of chlorine (Cl2). Using the information in the equation, explain how you would determine which reactant is in excess and calculate the maximum mass of aluminium chloride (AlCl3) that could be formed.
(Ar: Al = 27, Cl = 35.5)
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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Must show: method for finding moles, use chemical equation ratio, identify limiting reactant, calculate maximum product mass
Common mistakes: not identifying limiting reactant; mass not linked to correct reactant
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Calculates moles of both reactants (Al: 27/27 = 1 mol, Cl2: 85.0/71 = 1.2 mol)
M2 · 1 mark · Identifies Al as limiting reactant by mole ratio in equation (2 mol Al reacts with 3 mol Cl2) (follow-through)
A1 · 1 mark · Calculates maximum mass of AlCl3 formed by using limiting reactant: 1 mol Al forms 1 mol AlCl3 (133 g); answer 133 g (accept 133 – 134 g with working) (follow-through)
Worked solution
- Calculate moles of each reactant:
- According to the equation, 2 mol Al reacts with 3 mol Cl2.
- 1.0 mol Al would need 1.5 mol Cl2, but only 1.2 mol available, so Cl2 is actually the limiting reactant.
- But let's check: To react with 1.2 mol Cl2, we need (2/3)*1.2 = 0.8 mol Al (so Al is in excess). Maximum product depends on Cl2.