P = 100 kPa
T = 25oC + 273 = 298 K
V = 500 mL = 0.5 L
R = 8.314 kPa L K-1 mol-1
Solution:
The balanced chemical equation:
6HCl(aq) + 2Al(s) = 2AlCl3(aq) + 3H2(g)
According to the chemical equation: n(Al)/2 = n(H2)/3
The ideal gas law can be used to calculate the amount of H2:
PV = nRT
n= PV/RT
Thus:
n(H2) = (100 kPa × 0.5 L) / (8.314 kPa L K-1 mol-1 × 298 K) = 0.02 mol
n(Al)/2 = n(H2)/3
n(Al) = 2 × n(H2) / 3 = 2 × (0.02 mol) / 3 = 0.0133 mol
Moles of Al = Mass of Al / Molar mass of Al
The molar mass of Al is 27 g/mol
Thus:
Mass of Al = n(Al) × M(Al) = (0.0133 mol) × (27 g/mol) = 0.3591 g = 0.36 g
Answer: 0.36 g of aluminum (Al) is needed.
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