Question

amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted?
...A few of the molar masses are as follows:

Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.

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

Answer 1

We're giving quantities of both reactants, so we first need to find the limiting reagent. However, the question already asks for how much Al2S3 remains, so we know it's in excess and H2O is limiting. We can figure out how much Al2S3 is left by first converting to moles:

Al2S3: 20 g / 150.17 g/mol = 0.133 mol
H2O: 2 g / 18.02 g/mol = 0.111 mol

We know the reaction proceeds in a 1 Al2S3 : 6 H2O ratio, and we also know all 0.111 mol of H2O reacted. Now we can solve for the moles of Al2S3 reacted:

\[\frac{x \space mol \space Al_2S_3}{0.111 \space mol \space H_2O}=\frac{1 \space mol \space Al_2S_3}{6 \space mol \space H_2O}\]
\[x = 0.0185\]
So 0.0185 mol of Al2S3 reacted. This leaves 0.133-0.0185 = 0.1145 mol of Al2S3 left. Convert this back to mass by multiplying by molar mass: 0.1145 mol x 150.17 g/mol =17.19 g.

17.19 g of Al2S3 remains after the reaction! Let me know if you have any questions.

Answer 2

@matt101
Thanks