Hydrogen sulfide gas (H₂S, 34.08 g/mol) can be produced from the reaction of aluminum sulfide (150.2 g/mol) and water (18.02 g/mol) according to the following reaction equation: AL₂ S₃ (s) + 6 H₂ O(l) $\rightarrow$ 3 H₂ S(g) + 2 Al(OH) ₃ (s)
If 65.0 g of Al₂S₃ reacts with excess H₂O at 90.0 $\degree$ C and 0.987 atm, what volume of H₂S gas will be produced?

Question

Hydrogen sulfide gas (H₂S, 34.08 g/mol) can be produced from the reaction of aluminum sulfide (150.2 g/mol) and water (18.02 g/mol) according to the following reaction equation: AL₂ S₃ (s) + 6 H₂ O(l)

\rightarrow

3 H₂ S(g) + 2 Al(OH) ₃ (s)
If 65.0 g of Al₂S₃ reacts with excess H₂O at 90.0

\degree

C and 0.987 atm, what volume of H₂S gas will be produced?

Quizplus · Accepted Answer

First, calculate the moles of Al2S3 used: $65.0 \, \text{g} \, Al_2S_3 \times \frac{1 \, \text{mol} \, Al_2S_3}{150.2 \, \text{g}} = 0.433 \, \text{mol} \, Al_2S_3$. From the reaction, 3 moles of H2S are produced for every mole of Al2S3, so $0.433 \, \text{mol} \, Al_2S_3 \times \frac{3 \, \text{mol} \, H2S}{1 \, \text{mol} \, Al_2S_3} = 1.30 \, \text{mol} \, H2S$. Using the ideal gas law PV = nRT and converting temperature to Kelvin (90.0°C = 363.15 K), the volume of H2S can be calculated as $V = \frac{nRT}{P} = \frac{1.30 \, \text{mol} \times 0.0821 \, \text{L·atm/mol·K} \times 363.15 \, \text{K}}{0.987 \, \text{atm}} = 39.2 \, \text{L}$.

Hydrogen Sulfide Gas (H2S, 34 →\rightarrow→ 3 H2 S(g) + 2 Al(OH)3 (S) If 65

Hydrogen Sulfide Gas (H₂S, 34 $\rightarrow$ 3 H₂ S(g) + 2 Al(OH)₃ (S)
If 65