Final answer:
Using stoichiometry, we determine Fe2O3 as the limiting reactant and calculate that 75.0 g of Fe2O3 will produce 25.4 g of water when reacting with 4.50 g of H2 (rounded to three significant figures).
Step-by-step explanation:
To calculate the grams of water produced from the reaction of 75.0 g Fe2O3 and 4.50 g H2, we first need to determine the limiting reactant using stoichiometry and the balanced chemical equation:
3H2 + Fe2O3 → 2Fe + 3H2O
The molar mass of Fe2O3 is 159.7 g/mol, and the molar mass of H2 is 2.02 g/mol. Starting with Fe2O3:
75.0 g Fe2O3 × (1 mol Fe2O3 / 159.7 g Fe2O3) = 0.470 mol Fe2O3
For H2:
4.50 g H2 × (1 mol H2 / 2.02 g H2) = 2.23 mol H2
Next, using stoichiometry, we determine the amount of water produced from the limiting reactant. Since Fe2O3 is the limiting reactant, the moles of water produced will be:
0.470 mol Fe2O3 × (3 mol H2O / 1 mol Fe2O3) = 1.41 mol H2O
The molar mass of H2O is 18.02 g/mol. Therefore, the grams of water produced is:
1.41 mol H2O × (18.02 g H2O / 1 mol H2O) = 25.4 g H2O, rounded to three significant figures.