# Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 73

We can obtain 699g of $Fe$, but, the percent yield was only 93.5%.

#### Work Step by Step

1. Calculate the number of moles of $Fe_2O_3$: 55.85* 2 + 16* 3 = 159.7g/mol $1000g \times \frac{1 mol}{ 159.7g} = 6.26mol (Fe_2O_3)$ - The balanced reaction is: $Fe_2O_3 + 3CO -- \gt 2Fe + 3CO_2$ The ratio of $Fe_2O_3$ to $Fe$ is 1 to 2: $6.26 mol (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 12.5mol (Fe)$ 2. Calculate the mass of $Fe$: 55.85* 1 = 55.85g/mol $12.5 mol \times \frac{ 55.85 g}{ 1 mol} = 699g (Fe)$ 3. Now, calculate the percent yield: $\frac{ 654g}{699g} \times 100\% = 93.5\%$

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