Question

the reaction of iron ore with carbon follows the equation: 2fe2o3 + 3c → 4fe + 3co2.
in a reaction, 500 g of fe2o3 are mixed with 100 g of c. the actual yield of fe is 280.0 g, what is the percent yield of iron?
40%
20%
10%
80%
60%

Explanation:

Step1: Calculate the molar mass of \(Fe_2O_3\) and \(Fe\)

The molar mass of \(Fe_2O_3\): \(M_{Fe_2O_3}=(2\times55.85)+(3\times16)=159.7\space g/mol\)
The molar mass of \(Fe\): \(M_{Fe} = 55.85\space g/mol\)

Step2: Determine the limiting reactant

Moles of \(Fe_2O_3\): \(n_{Fe_2O_3}=\frac{500\space g}{159.7\space g/mol}\approx3.13\space mol\)
Moles of \(C\): \(n_{C}=\frac{100\space g}{12\space g/mol}\approx8.33\space mol\)
From the balanced equation \(2Fe_2O_3 + 3C
ightarrow4Fe+3CO_2\), the mole ratio of \(Fe_2O_3\) to \(C\) is \(2:3\).
For \(3.13\space mol\) of \(Fe_2O_3\), the moles of \(C\) required is \(n_{C_{required}}=\frac{3}{2}\times3.13 = 4.695\space mol\lt8.33\space mol\). So \(Fe_2O_3\) is the limiting reactant.

Step3: Calculate the theoretical yield of \(Fe\)

From the balanced equation, \(2\space mol\) of \(Fe_2O_3\) produces \(4\space mol\) of \(Fe\).
For \(3.13\space mol\) of \(Fe_2O_3\), moles of \(Fe\) produced (theoretical) \(n_{Fe_{theoretical}}=\frac{4}{2}\times3.13 = 6.26\space mol\)
Theoretical mass of \(Fe\): \(m_{Fe_{theoretical}}=6.26\space mol\times55.85\space g/mol\approx350\space g\)

Step4: Calculate the percent yield

Percent yield \(=\frac{actual\space yield}{theoretical\space yield}\times100\%=\frac{280\space g}{350\space g}\times100\% = 80\%\)

Answer:

80%