## Chemistry: Molecular Approach (4th Edition)

- The limiting reactant is $CO$. - The theoretical yield is equal to 114 g Fe. - The percent yield is 63.4%.
1. Calculate the number of moles of $Fe_2O_3$: 55.85* 2 + 16* 3 ) = 159.7g/mol $167g \times \frac{1 mol}{ 159.7g} = 1.046mol (Fe_2O_3)$ According to the balanced equation: The ratio of $Fe_2O_3$ to $Fe$ is 1 to 2: $1.046 mol (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 2.091mol (Fe)$ ------------------------------ 3. Calculate the number of moles of $CO$: 12.01* 1 + 16* 1 = 28.01g/mol $85.8g \times \frac{1 mol}{ 28.01g} = 3.063mol (CO)$ According to the balanced equation: The ratio of $CO$ to $Fe$ is 3 to 2: $3.063 mol (CO) \times \frac{ 2 mol(Fe)}{ 3 mol (CO)} = 2.042mol (Fe)$ ------------------------ Therefore, $CO$ is the limiting reactant, so we should use the calculations where $2.042$ mol Fe is produced. 4. Calculate the mass of $Fe$: 55.85* 1 = 55.85g/mol $2.042 mol \times \frac{ 55.85 g}{ 1 mol} = 114g (Fe)$ 5. Now, calculate the percent yield: $\frac{72.3g}{114g} \times 100\% = 63.4\%$