Combining 0288 mol of Fe O with excess carbon produced 110 g

Combining 0.288 mol of Fe O, with excess carbon produced 11.0 g of Fe. Fe,03 +3C 2Fe +3CO What is the actual yield of iron in moles? Number mol What was the theoretical yield of iron in moles? Number mol What was the percent yield? Number

Solution

From the balanced equation

1 moles of Fe2O3 gives 2 moles of Fe

Hence 0.288mol of Fe2O3 gives 2x0.288mol Fe

moles of Fe theoretically = 2 x0.288 mol

mass of Fe to obtained theoretically = 2x0.288 x 55.845 g/mol

=32.17 g

1) The actual yield in moles = mass/ atomic weight

= 11.0g/55.845

= 0.1969 moles

2) The thoretical yiled of iron = 32.17 g

= 0.576 moles

3)% yield = actual yield x100/ theoretica yield

= 0.1969x100/0.576

= 34.19%