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