2Al + 6HCl → 2AlCl3 + 3H2
M(Al) = 27.0 g/mol
n(Al) "= \\frac{27}{27} = 1 \\;mol"
According to the reaction:
n(H2) "= \\frac{3}{2}n(Al) = \\frac{3}{2} \\times 1 = 1.5 \\;mol"
M(H2) = 2 g/mol
m(H2) "= 1.5 \\times 2 = 3 \\;g"
Answer: 3 g
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