Question

If 55.5 g of molten iron(III) oxide reacts with 30.1 g of aluminum, what is the mass of iron produced? Fe2O3(l)+Al(l)⟶ΔFe(l)+Al2O3(s) Express your answer with the appropriate units.

Answer 1

Molar mass of Fe2O3,

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

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass(Fe2O3)= 55.5 g

use:

number of mol of Fe2O3,

n = mass of Fe2O3/molar mass of Fe2O3

=(55.5 g)/(1.597*10^2 g/mol)

= 0.3475 mol

Molar mass of Al = 26.98 g/mol

mass(Al)= 30.1 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(30.1 g)/(26.98 g/mol)

= 1.116 mol

Balanced chemical equation is:

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

1 mol of Fe2O3 reacts with 2 mol of Al

for 0.3475 mol of Fe2O3, 0.6951 mol of Al is required

But we have 1.116 mol of Al

so, Fe2O3 is limiting reagent

we will use Fe2O3 in further calculation

Molar mass of Fe = 55.85 g/mol

According to balanced equation

mol of Fe formed = (2/1)* moles of Fe2O3

= (2/1)*0.3475

= 0.6951 mol

use:

mass of Fe = number of mol * molar mass

= 0.6951*55.85

= 38.82 g

Answer: 38.8 g