Determine the mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \). Express your answer in grams of iron to three significant figures.

To determine the mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \), we need to calculate the molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) and then use the molar mass to find the mass of iron. The molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) can be calculated as follows: - The molar mass of iron (Fe) is approximately 55.845 g/mol. - The molar mass of oxygen (O) is approximately 16.00 g/mol. The molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) is: \[ 2 \times 55.845 + 3 \times 16.00 = 111.69 + 48.00 = 159.69 \, \text{g/mol} \] Now, we can calculate the mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) using the molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) and the molar mass of iron. The mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) is: \[ \text{Mass of iron} = \frac{2 \times 55.845}{159.69} \times 79.2 \] Let's calculate this. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(2\times 55.845\right)}{159.69}\times 79.2\) - step1: Remove the parentheses: \(\frac{2\times 55.845}{159.69}\times 79.2\) - step2: Multiply the numbers: \(\frac{111.69}{159.69}\times 79.2\) - step3: Divide the terms: \(\frac{3723}{5323}\times 79.2\) - step4: Convert the expressions: \(\frac{3723}{5323}\times \frac{396}{5}\) - step5: Multiply the numbers: \(\frac{1474308}{26615}\) The mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) is approximately 55.393 g. Therefore, the mass of iron in 79.2 g of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) is 55.393 g.