Question

how many grams of al are needed to produce 0.450 l of 1.6 m alcl₃?
2al + 6hcl → 2alcl₃ + 3h₂
? g al

Explanation:

Step1: Calculate moles of \( \text{AlCl}_3 \)

Molarity formula: \( M = \frac{n}{V} \), so \( n = M \times V \).
Given \( M = 1.6 \, \text{M} \), \( V = 0.450 \, \text{L} \),
\( n_{\text{AlCl}_3} = 1.6 \, \text{mol/L} \times 0.450 \, \text{L} = 0.72 \, \text{mol} \).

Step2: Relate moles of \( \text{Al} \) and \( \text{AlCl}_3 \)

From reaction \( 2\text{Al}
ightarrow 2\text{AlCl}_3 \), mole ratio \( \text{Al} : \text{AlCl}_3 = 1:1 \).
Thus, \( n_{\text{Al}} = n_{\text{AlCl}_3} = 0.72 \, \text{mol} \).

Step3: Calculate mass of \( \text{Al} \)

Molar mass of \( \text{Al} = 26.98 \, \text{g/mol} \).
Mass formula: \( m = n \times M \),
\( m_{\text{Al}} = 0.72 \, \text{mol} \times 26.98 \, \text{g/mol} \approx 19.43 \, \text{g} \).

Answer:

\( \approx 19.4 \) (or 19.43) grams