Question

How many grams of Chlorine will react completely with 32.9 g of Aluminum?

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Answer 1

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