Final answer:
When 21.1 g of aluminum reacts with excess HCl, 1.21 moles of H₂ gas will be produced. This is found by using stoichiometry to convert grams of Al to moles and then using the balanced equation to find the moles of H₂.
Step-by-step explanation:
The question asks if 21.1 g of aluminum (Al) react with excess HCl, how many moles of H₂ gas will be produced. To solve this, we need to use the molar mass of aluminum and stoichiometry from the balanced chemical equation of the reaction between aluminum and hydrochloric acid.
The balanced chemical equation is:
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)
First, we calculate the number of moles of Al:
21.1 g Al × (1 mol Al / 26.98 g Al) = 0.782 moles Al
From the stoichiometry of the balanced equation, we know that 2 moles of Al produce 3 moles of H₂. Therefore, we can calculate the moles of H₂ produced as follows:
0.782 moles Al × (3 moles H₂ / 2 moles Al) = 1.173 moles H₂
1.173 moles of H₂ gas rounds to 1.21 moles when considering significant figures, so the answer is 1.21 moles of H₂ gas, which corresponds to option (b).