Final answer:
The volume of hydrogen gas produced at STP from the reaction of 37.2 g of Al with HCl is 46.30 liters, found by first converting mass to moles of Al, then using the mole ratio from the reaction, and finally applying the ideal gas law at STP.
Step-by-step explanation:
The subject of this question is the calculation of the volume of hydrogen gas produced from the reaction of aluminum with hydrochloric acid. The first step is to use the molar mass of aluminum (26.98 g/mol) to convert the mass of aluminum given (37.2 g) into moles. One mole of Al will produce 1.5 moles of H2 according to the balanced chemical equation (2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)). Next, we apply the ideal gas law at standard temperature and pressure (STP, defined as 0°C and 1 atm) where one mole of any gas occupies 22.4 liters.
Firstly, calculate the moles of Al:
37.2 g Al × (1 mol Al / 26.98 g Al) = 1.378 moles of Al
Since the ratio of Al to H2 in the chemical equation is 2:3, we will have (1.378 mol Al) × (3 mol H2 / 2 mol Al) = 2.067 moles of H2.
Finally, convert moles of H2 to liters:
2.067 mol H2 × 22.4 L/mol = 46.30 L H2
Therefore, the volume of hydrogen gas produced at STP by the reaction of 37.2 g of Al with excess HCl is 46.30 liters.