Question

The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals.
(Atomic mass of Al = 27 u Atomic mass of O = 16 u)
\[\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}\]; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.

1. Calculate the mass of Al₂O₃ formed.
2. How much of the excess reagent is left at the end of the reaction?

Answer 1

Given \[\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}\]

	Reactants		Products
	Al	Fe2O3	Al2O3	Fe
Amount of reactant allowed to react	324 g	1.12 kg	-	-
Number of moles allowed to react	`324/27` = 12 mol	`(1.12 xx 10^3)/160` = 7 mol	-	-
Stoichiometric Co-efficient	2	1	1	2
Number of moles consumed during the reaction	12 mol	6 mol	-	-
Number of moles of reactant unreacted and number of moles of product formed	-	1 mol	6 mol	12 mol

Molar mass of Al₂O₃ formed = 6 mol × 102 g mol^–1 ........[Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102]

= 612 g

Excess reagent = Fe₂O₃.
Amount of excess reagent left at the end of the reaction = 1 mol × 160 g mol^–1

= 160 g ........[Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160 g]

= 160 g