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Question

5 mol of aluminium chloride is mixed with 280 g of magnesium oxide, to give aluminium oxide. Find the amount of the excess reagent left after the reaction.
2AlCl3+3MgOAl2O3+3MgCl2

A
44.055 g
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B
42.032 g
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C
54.056 g
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D
32.056 g
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Solution

The correct option is A 44.055 g
2AlCl3+3MgOAl2O3+3MgCl2
Moles of MgO=28040=7 mol of MgO
2 mol of AlCl3 reacts with 3 mol of MgO
5 mol will react with = 7.5 mol
So, MgO is the limiting reactant.
3 mol of MgO requires 2 mol of AlCl3
7 mol of MgO will require = 4.67 mol
Excess reagent left = 5 – 4.67 = 0.33 mol
Mass of AlCl3 left =0.33×133.5=44.055 g

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