Answer to Question #252531 in General Chemistry for Rodel

Question

Iron can be obtained by reacting 200g of Fe₂O₃ with 42g of CO.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

How much of the excess reagent in grams is left at the end of the reaction? And, what will be the percent yield if 48g of Fe is the actual yield?

Answer 1

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Mole ratio:-1:3:2:3

Moles of iron yielded $=\frac{mass}{RAM}$

$=\frac{48g}{56g}=0.8571moles$

Mole ratio of Fe₂O₃:Fe from the equation above is 1:2

$\therefore$ Moles of Fe₂O₃ reaction=$\frac{1}{2}$ ×Moles of Fe yielded.

=$\frac{1}{2}×0.8571moles$ =0.42855moles

But moles of Fe₂O₃ reacting =$\frac{mass reacting}{RFM of Fe_2 O_3}$

$0.42855=\frac{mass}{(56×2)+(16×3)}$

$\therefore$ Mass of Fe₂O₃ reacted =0.42855×160=68.568g

But mass of excess Fe₂O₃ =200g -68.568g =131.432g

Mass of excess Fe₂O₃=131.432g

%yield if 48g of Fe is the actual yield=$\frac{68.568g}{200g}×$ 100%=34.284%

=34.284%