Combining 0.281 mol Fe2O3 with excess carbon produced 14.2 g Fe. Fe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles?

Question

Combining 0.281 mol Fe2O3 with excess carbon produced 14.2 g Fe.

Fe2O3+3C⟶2Fe+3CO

What is the actual yield of iron in moles?

actual yield:__?__mol

What is the theoretical yield of iron in moles?

theoretical yield:___?___mol

What is the percent yield?

percent yield:___?___%

Answer 1

The actual yield of iron is 14.2 g

The theoretical yield is calculated as follows:

0.281 mols Fe₂O₃ x 2 mols Fe / mol Fe₂O₃ = 0.562 mols Fe theoretical yield

Percent yield = actual yield in g / theoretical yield in g (x100%)

actual = 14.2 g

theoretical = 0.562 g Fe x 55.9 g/mol = 31.4 g

% yield = 14.2 g / 31.4 g (x100%) = 45.2% yield