Question

The iron oxide $F{e}_2{O}_3$reacts with carbon monoxide to give iron and carbon dioxide

$F{e}_2{O}_3+3CO\to 2Fe+3C{O}_2$

The reaction of 433.2g$F{e}_2{O}_3$ with excess CO yields 254.3g iron. Calculate the theoretical yield of iron (assuming complete reaction) and its percentage yield.

Short answer

The theoretical yield of iron is303.0 g.

The percentage yield of Fe will be 83.97%.

Step-by-step solution

Process of calculation of theoretical yield

We are given with the yield of iron when a certain amount of $F{e}_2{O}_3$reacts with excess CO.

Iron oxide reacts with carbon monoxide to form iron and carbon dioxide. The equation for this reaction is

$F{e}_2{O}_3+3CO\to 2Fe+3C{O}_2$

Iron oxide reacts with carbon monoxide, so the limiting reactant is iron oxide and the theoretical yield is calculated using its moles.

Calculation of theoretical yield of iron

According to the balanced reaction 1 mole of $F{e}_2{O}_3$produces 2 moles of Fe. The mass of Fe produced from 433.2g$F{e}_2{O}_3$ is

$$\begin{gathered} =433.2F{e}_2{O}_3\times \frac{1molF{e}_2{O}_3}{159.7g.F{e}_2{O}_3}\times \frac{2moleFe}{1moleF{e}_2{O}_3}\times \frac{55.85G.Fe}{1mole.Fe} \\ =303.0g.Fe \end{gathered}$$

Therefore, theoretical yield of iron is 303.0 g.

 Calculation of % yield of Fe

The formula for percentage yield is

$Percentageyield=\frac{Actualyield}{Theoreticalyield}\times 100\%$

The ratio of the actual yield of iron is 254.3g.

$$\begin{gathered} Percentageyield=\frac{254.3g}{303.0g}\times 100\% \\ =83.927\% \end{gathered}$$

So, the % yield of Fe will be 83.97%.