DATA GIVEN:

Mass = 15.00g of Al₂S₃
Mass of water = 10.00g
Asked:
a/ Limiting reagent
b/ Maximum mass of H₂S that can be formed
c/ Excess reagent remain after the reaction is complete

Chemical equation for the reaction.
Al₂S₃ + 6H₂O → 2Al(OH)₃ + 3H₂S

a/ Limiting reagent

1 mole: 6 moles: 2mole: 3mole
To convert 15g of Al₂S₃ into mole
Mole =mass /Molar mass of Al₂S₃
= 15g /150 g/mole
= 0.1

Mole ratio for Al₂S₃ = Mole /Mole coefficient
= 0.1 mole /1 mole
= 0.1 mole

Mole of water (H₂O) = Mass /Molar mass
= 10.00g /18g/ mole
= 0.56 mole

Mole ratio for water = 0.56 mole /6
= 0.56 mole

The limiting is water excess is Aluminium Sulphide.

b/ Maximum mass of H₂S formed

Mole equivalent between water and hydrogen Sulphide.
6 mole 0f H2O Ξ 3 mole of H₂S
0.56 mole of H₂O Ξ ??
=0.56 mole × 3 mole /6 mole
= 0.28 mole of H₂S

Mass = mole × molar mass of H₂sS
= 0.28 mole × 32g/mole
= 8.96g

8.96g of hydrogen Sulphide can be formed.

c/ Mass of excess reagent remains

from;
1 mole of Al₂S₃ Ξ 6 mole of H₂O
? Ξ 0.56 mole of H₂O

1mole × 0.56 mole = X x 6 mole
X = 1 mole × 0.56 mole /6mole
= 0.00933 mole of Al₂S₃

To convert 0.0933 mole into mass is 13.995g.

Mass remain = Original mass – mass reacted
= 15g -13.995g
= 1.005g

The excess reagent remain of Al₂S₃ was found to be 1.005