Question

15.00 g of aluminum sulfide (150.1 g/mol) and 10.00 g of water (18.02 g/mol) react until the limiting reactant is used up. Calculate the mass of H2S (34.08 g/mol) that can be produced from these reactants. You will need to balance the reaction equation.

Answer 1

10.224g

Step-by-step explanation:

Al2S3(s) + 6H2O(l) =>>> 2Al(OH)3(s) + 3H2S(g).

aluminum sulphide is the limiting reactant

if 1 moles of aluminum sulphide reacts to give 3 moles of H2S

then 0.1 moles of aluminum sulphide will give x

3 ×0.1/1 = 0.3

to get the mass, Nm =mass/ molar mass

mass= 0.3 × 34.08

=10.224g