How many grams of Chlorine will react completely with 32.9 g of Aluminum?
2Al + 3Cl2 → 2AlCl3
MM(Al) = 26.98 g/mol
n(Al) "= \\frac{32.9}{26.98} = 1.22 \\;mol"
According to the reaction:
n(Cl2) "= \\frac{3}{2}n(Al) = \\frac{3}{2} \\times 1.22 = 1.83 \\;mol"
MM(Cl2) = 70.9 g/mol
m(Cl2) "= 1.83 \\times 70.9 = 129.75 \\;g"
Answer: 129.75 g
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