Question

If 150.93 grams of aluminum metal is reacted with 19.26grams of iron (III) oxide, what mass of excess reactant will be left over after the reaction?Reaction: 2AI(s) + Fe2O3(aq) --> AI2O3(aq) =2Fe(s)

Answer 1

143.49 g Al

Procedure

Using the following balanced equation determine the limiting reagent.

2AI(s) + Fe₂O₃(aq) --> AI₂O₃(aq) + 2Fe(s)

First, convert all the reagents to moles and then use the mole ratios method

`150.93\text{ g Al}\frac{1\text{ mol Al}}{26.98\text{ g Al}}=5.59\text{ }mol\text{ Al}`
`19.26\text{ g Fe}_2\text{O}_3\frac{1\text{ mol Fe}_2\text{O}_3}{159.68\text{ g Fe}_2\text{O}_3}=0.1206\text{ }mol\text{ Fe}_2\text{O}_3`

Molar ratios

`5.59\text{ mol Al}\frac{2\text{ mol Fe}}{2\text{ mol Al}}=5.59\text{ }mol\text{ Fe}_`
`0.1206\text{ mol Fe}_2\text{O}_3\frac{2\text{ Fe}}{\text{ 1 mol Fe}_2\text{O}_3}=0.2412\text{ Fe}_`

Therefore the limiting reagent is Fe₂O₃

Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess reagent given.

`19.26\text{ g Fe}_2\text{O}_3\frac{1\text{ mol Fe}_2\text{O}_3}{159.68\text{ g Fe}_2\text{O}_3}\frac{2\text{ mol Al }}{1\text{ mol Fe}_2O_3}\frac{26.98\text{ g Al}}{1\text{ mol Al}}=6.5085\text{ g Al}`

Excess = Mass of total excess reagent given – mass of excess reagent consumed in the reaction

`Excess=150.93-6.5085=143.49\text{ g Al}`