Given the reaction: Fe3O4 + 4 H2 → 3 Fe + 4 H2O, how many grams of Fe3O4 are required to react completely with 300 grams of H2?

• A. 1,195 g
• B. 800 g
• C. 2,400 g
• D. 3,600 g

Answer: (A)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{Fe_3O_4 + 4\ H_2 \rightarrow 3\ Fe + 4\ H_2O}\).
2. Calculate the molar mass of \(\mathrm{H_2}\): since hydrogen has an atomic mass of approximately 1 g/mol, \(\mathrm{H_2}\) has a molar mass of \$2\ \mathrm{g/mol}$.
3. Determine the number of moles of \(\mathrm{H_2}\) given 300 grams using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{300\ \mathrm{g}}{2\ \mathrm{g/mol}}\).
4. Use the stoichiometric ratio from the balanced equation to find moles of \(\mathrm{Fe_3O_4}\) needed: for every 4 moles of \(\mathrm{H_2}\), 1 mole of \(\mathrm{Fe_3O_4}\) reacts. So, \(\text{moles of } \mathrm{Fe_3O_4} = \frac{\text{moles of } \mathrm{H_2}}{4}\).
5. Calculate the molar mass of \(\mathrm{Fe_3O_4}\) by summing the atomic masses: iron (Fe) is about 55.85 g/mol and oxygen (O) is about 16.00 g/mol, so \(\mathrm{Fe_3O_4}\) molar mass = \$3 \times 55.85 + 4 \times 16.00\(. Then, find the mass of \)\mathrm{Fe_3O_4}$ required by multiplying its moles by its molar mass.