What is the 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 and al2s3(s) + 6 h2o(l) → 2 al(oh)3(s) + 3 h2s(g)

From the equation, we can tell that 1 mol of Al₂S₃ requires 6 moles of water.
The molar ratio is 1/6
Moles of Al₂S₃ present = 20/150.17
= 0.133
Moles of water present = 2/18.02
= 0.111
The moles of Al₂S₃ that will react are:
0.111/6
= 0.0185
The remaining amount:
0.133 - 0.0185
= 0.1145 mol
Or
0.1145 * 150.17
= 17.19 grams

Forgot the question hahaha