Question

Calculate the grams of the indicated product when 17.9 g of the first reactant and 10.1 g of the second reactant is used:
Al2S3(s)+6H2O(l)→2Al(OH)3(aq)+3H2S(g) (H2S)

Answer 1

Al₂S₃(s) + 6H₂O(l) → 2Al(OH)₃(aq) + 3H₂S(g)

Calculating amounts of reactants:

$n(Al_2S_3) = \frac{m(Al_2S_3)}{M(Al_2S_3)} = \frac{17.9g}{(2\cdot 27 + 3 \cdot 32)\frac{g}{mol}}=\frac{17.9g}{150\frac{g}{mol}}=0.119mol$

$n(H_2O) = \frac{m(H_2O)}{M(H_2O)} = \frac{10.1g}{(2\cdot 1 + 16)\frac{g}{mol}}=\frac{10.1g}{18\frac{g}{mol}}=0.561mol$

According to stoichiometry of the reaction ratio Al₂S₃:H₂O is 1:6. But according to given amounts of reagents this ratio is 0.119:0.561 = 1:4.7. This means that Al₂S₃ is excess reactant and H₂O is limiting reactant.

Amount of Al₂S₃ reacted with water is:

$n(react. Al_2S_3)=\frac{1}{6}n(H_2O)=\frac{1}{6}\cdot 0.561mol = 0.0935 mol$

Amount of Al(OH)₃ produced in the reaction:

$n(Al(OH)_3)=2n(react. Al_2S_3)=2\cdot 0.0935 mol = 0.187 mol$

$m(Al(OH)_3)=n(Al(OH)_3)\cdot M(Al(OH)_3) = 0.187mol \cdot 78 \frac{g}{mol}=14.6g$

Amount of H₂S produced in the reaction:

$n(H_2S)=3n(react. Al_2S_3)=3\cdot 0.0935 mol = 0.2805mol$

$m(H_2S)=n(H_2S)\cdot M(H_2S) = 0.2805mol \cdot 34 \frac{g}{mol}=9.5g$