Question

Combining 0.221 mol of Fe2O3 with excess carbon produced 10.6 g of Fe.

What is the actual yield of iron in moles?

What was the theoretical yield of iron in moles?

What was the percent yield?

Answer 1

The balanced equation is

Fe2O3 + 3 C -----> 2 Fe + 3 CO

Number of moles of Fe = 10.6 g / 55.845 g/mol = 0.190 mole

Therefore, the actual yield of iron = 0.190 mole

From the balanced equation we can say that

1 mole of Fe2O3 produces 2 mole of Fe so

0.221 mole of Fe2O3 will produce

= 0.221 mole of Fe2O3 *(2 mole of Fe / 1 mole of Fe2O3)

= 0.442 mole of Fe

Therefore, theoretical yield of Fe = 0.442 mole

percent yield = (actual yield / theoretical yield)*100

percent yield = (0.190 / 0.442)*100

percent yield = 43.0