Chemistry 10th Edition

In this reaction, 87.22 g of $Fe_3O_4$ reacted.
1. Calculate the number of moles of $H_2O$: 1.008* 2 + 16.00* 1 = 18.02g/mol $27.15g \times \frac{1 mol}{ 18.02g} = 1.507mol (H_2O)$ According to the balanced equation: The ratio of $H_2O$ to $Fe_3O_4$ is 4 to 1: $1.507 mol (H_2O) \times \frac{ 1 mol(Fe_3O_4)}{ 4 mol (H_2O)} = 0.3767mol (Fe_3O_4)$ 2. Calculate the mass of $Fe_3O_4$: 55.85* 3 + 16.00* 4 = 231.55g/mol $0.3767 mol \times \frac{ 231.55 g}{ 1 mol} = 87.22g (Fe_3O_4)$