Question

If 65g of Aluminum reacts with excess Chlorine, how many grams of Aluminum Chloride will form?

Answer 1

2Al + 3Cl₂ → 2AlCl₃

MM(Al) = 26.98 g/mol

n(Al) $= \frac{65}{26.98} = 2.41 \;mol$

According to the reaction:

n(AlCl₃) =n(Al) = 2.41 mol

MM(AlCl₃) = 133.34 g/mol

m(AlCl₃) $= 2.41 \times 133.34 = 321.35 \;g$