QuestionMarch 21, 2026

What mass of HCl is needed to generate 452 g of AlCl_(3) ? 2Al+6HClarrow 2AlCl_(3)+3H_(2) AlCl_(3):133.33g/mol HCl:36.46g/mol [?]gHCl

What mass of HCl is needed to generate 452 g of AlCl_(3) ? 2Al+6HClarrow 2AlCl_(3)+3H_(2) AlCl_(3):133.33g/mol HCl:36.46g/mol [?]gHCl
What mass of HCl is needed to generate 452 g of AlCl_(3) ? 2Al+6HClarrow 2AlCl_(3)+3H_(2) AlCl_(3):133.33g/mol HCl:36.46g/mol [?]gHCl

Solution
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Answer

37.05 g Explanation 1. Calculate moles of AlCl_{3} n = \frac{m}{M} = \frac{45.2}{133.33} \approx 0.339 \ \text{mol} 2. Determine moles of HCl required Reaction: 2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2} Ratio: 2 \ \text{mol} \ AlCl_{3} per 6 \ \text{mol} \ HCl → 1 \ \text{mol} \ AlCl_{3} needs 3 \ \text{mol} \ HCl n_{HCl} = 0.339 \times 3 \approx 1.017 \ \text{mol} 3. Convert moles of HCl to mass m = n \times M = 1.017 \times 36.46 \approx 37.05 \ \text{g}

Explanation

1. Calculate moles of $AlCl_{3}$ <br /> $n = \frac{m}{M} = \frac{45.2}{133.33} \approx 0.339 \ \text{mol}$ <br /><br />2. Determine moles of HCl required <br /> Reaction: $2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}$ <br /> Ratio: $2 \ \text{mol} \ AlCl_{3}$ per $6 \ \text{mol} \ HCl$ → $1 \ \text{mol} \ AlCl_{3}$ needs $3 \ \text{mol} \ HCl$ <br /> $n_{HCl} = 0.339 \times 3 \approx 1.017 \ \text{mol}$ <br /><br />3. Convert moles of HCl to mass <br /> $m = n \times M = 1.017 \times 36.46 \approx 37.05 \ \text{g}$
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