If the reaction yield is 94.4%, what mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 g of water? The balanced chemical equation is: Mg (s) + 2 H2O (l) → Mg(OH)2 (s) + H2 (g).

Step 3: Determine the limiting reactant. Calculate the moles of magnesium and water available: \( \text{moles of Mg} = \frac{4.73 \text{ g}}{24.31 \text{ g/mol}} \) and \( \text{moles of H}_2\text{O} = \frac{1.83 \text{ g}}{18.02 \text{ g/mol}} \). Compare the mole ratio from the balanced equation to find the limiting reactant.

Step 4: Use the stoichiometry of the balanced equation to calculate the theoretical yield of hydrogen gas. Based on the limiting reactant, use the mole ratio from the balanced equation to find the moles of hydrogen gas produced. Then convert moles of hydrogen gas to grams using its molar mass: \( \text{grams of H}_2 = \text{moles of H}_2 \times 2.02 \text{ g/mol} \).