Final answer:
To find the volume of hydrogen gas produced when 9.92 g of Al reacts with HCl at STP, we calculate the moles of Al, use the molar ratio to find moles of H2, and then convert moles to volume using the molar volume at STP. The closest volume produced is 24.7 L.
Explanation: the volume of hydrTo determineogen gas produced at Standard Temperature and Pressure (STP) when 9.92 g of aluminum (Al) reacts with hydrochloric acid (HCl), we need to use stoichiometry based on the balanced chemical equation provided:
6HCl(aq) + 2Al(s) → 2AlCl₃(aq) + 3H₂(g)
Firstly, calculate the number of moles of aluminum using its molar mass (26.98 g/mol):
moles of Al = mass (g) / molar mass (g/mol) = 9.92 g / 26.98 g/mol
Next, use the molar ratio from the balanced equation to find the moles of hydrogen gas.
The molar ratio of Al to H₂ is 2:3, so:
moles of H₂ = (3/2) * moles of Al
At STP, 1 mole of gas occupies 22.4 liters. Therefore, the volume of H₂ is:
volume of H₂ (L) = moles of H₂ * 22.4 L/mol
After calculations, the answer closest to the volume of H₂ produced is 24.7 L, which is choice b. Hence, reacting 9.92 g of Al with HCl at STP will yield approximately 24.7 L of hydrogen gas.