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"
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