Question

For the following reaction, 55.7 grams of iron(III) oxide are allowed to react with 23.5 grams of aluminum. iron(III) oxide(s) + aluminum(s) — aluminum oxide(s) + iron(s) What is the maximum amount of aluminum oxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1)

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass(Fe2O3)= 55.7 g

use:

number of mol of Fe2O3,

n = mass of Fe2O3/molar mass of Fe2O3

=(55.7 g)/(1.597*10^2 g/mol)

= 0.3488 mol

Molar mass of Al = 26.98 g/mol

mass(Al)= 23.5 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(23.5 g)/(26.98 g/mol)

= 0.871 mol

Balanced chemical equation is:

Fe2O3 + 2 Al ---> Al2O3 + 2 Fe

1 mol of Fe2O3 reacts with 2 mol of Al

for 0.3488 mol of Fe2O3, 0.6976 mol of Al is required

But we have 0.871 mol of Al

so, Fe2O3 is limiting reagent

we will use Fe2O3 in further calculation

Molar mass of Al2O3,

MM = 2*MM(Al) + 3*MM(O)

= 2*26.98 + 3*16.0

= 101.96 g/mol

According to balanced equation

mol of Al2O3 formed = (1/1)* moles of Fe2O3

= (1/1)*0.3488

= 0.3488 mol

use:

mass of Al2O3 = number of mol * molar mass

= 0.3488*1.02*10^2

= 35.56 g

Answer: 35.6 g

2)

Fe2O3 is limiting reagent

Answer: Fe2O3

3)

According to balanced equation

mol of Al reacted = (2/1)* moles of Fe2O3

= (2/1)*0.3488

= 0.6976 mol

mol of Al remaining = mol initially present - mol reacted

mol of Al remaining = 0.871 - 0.6976

mol of Al remaining = 0.1735 mol

Molar mass of Al = 26.98 g/mol

use:

mass of Al,

m = number of mol * molar mass

= 0.1735 mol * 26.98 g/mol

= 4.68 g

Answer: 4.68 g