Question

What is the theoretical yield of Al when 15 g of Al203 reacts with 15 g of Fe?

Answer 1

Molar mass of Al2O3,

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

= 2*27 + 3*16.0

= 102 g/mol

mass(Al2O3)= 15 g

number of mol of Al2O3,

n = mass/molar mass

=(15 g)/(102 g/mol)

= 0.147 mol

Molar mass of Fe = 56 g/mol

mass(Fe)= 15 g

number of mol of Fe,

n = mass/molar mass

=(15.0 g)/(56 g/mol)

= 0.268 mol

Balanced chemical equation is:

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

1 mol of Al2O3 reacts with 2 mol of Fe

for 0.147 mol of Al2O3, 0.294 mol of Fe is required

But we have 0.268 mol of Fe

so, Fe is limiting reagent

we will use Fe in further calculation

According to balanced equation

mol of Al formed = moles of Fe

= 0.268 mol

mass of Al = number of mol * molar mass

= 0.268*27

= 7.24 g

Answer: 7.24 g