Question

if 30.7 g of iron(iii) oxide is obtained from the reaction of 85.3 g of iron(iv) sulfide with excess oxygen, what is the percentage yield?
4 fes₂(s) + 11 o₂(g) → 2 fe₂o₃(s) + 8 so₂(g)
68.3 %
45.2 %
54.0 %
88.4 %
34.0 %

Explanation:

Step1: Calculate moles of \(FeS_2\)

Molar mass of \(FeS_2\) is \(M = 55.85+(2\times32.07)= 119.99\space g/mol\).
Moles of \(FeS_2\), \(n=\frac{m}{M}=\frac{85.3\space g}{119.99\space g/mol}\approx0.7117\space mol\)

Step2: Calculate theoretical moles of \(Fe_2O_3\)

From the balanced equation \(4FeS_2 + 11O_2
ightarrow2Fe_2O_3+8SO_2\), mole ratio of \(FeS_2:Fe_2O_3 = 4:2 = 2:1\)
Theoretical moles of \(Fe_2O_3\), \(n_{theo}=\frac{0.7117\space mol}{2}=0.35585\space mol\)

Step3: Calculate theoretical mass of \(Fe_2O_3\)

Molar mass of \(Fe_2O_3\) is \(M = 2\times55.85+(3\times16)=159.7\space g/mol\)
Theoretical mass of \(Fe_2O_3\), \(m_{theo}=n_{theo}\times M=0.35585\space mol\times159.7\space g/mol\approx56.83\space g\)

Step4: Calculate percentage yield

Percentage yield \(=\frac{actual\space yield}{theoretical\space yield}\times100\%=\frac{30.7\space g}{56.83\space g}\times100\%\approx54.0\%\)

Answer:

54.0 %