Answer
- The limiting reactant is $CO$.
- The theoretical yield is equal to 114 g Fe.
- The percent yield is 63.4%.
Work Step by Step
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:
$1.046 mol (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 2.091mol (Fe)$
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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:
$3.063 mol (CO) \times \frac{ 2 mol(Fe)}{ 3 mol (CO)} = 2.042mol (Fe)$
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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:
$\frac{72.3g}{114g} \times 100\% = 63.4\%$
