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. Notice that you will need to balance the reaction equation.

___Al2S3(s)+ ___H2O > ___Al(OH)3(s)+ ___H2S(g)

13.89 g d. 9.456 g

10.21 g e. 1.108 g

19.67 g

Answer 1

mass(Al2S3)= 15.0 g

number of mol of Al2S3,

n = mass/molar mass

=(15.0 g)/(150.1 g/mol)

= 0.100 mol

mass(H2O)= 10.0 g

number of mol of H2O,

n = mass/molar mass

=(10.0 g)/(18.02 g/mol)

= 0.555 mol

Balanced chemical equation is:

Al2S3 + 6 H2O ---> 3 H2S + 2 Al(OH)3

1 mol of Al2S3 reacts with 6 mol of H2O

for 0.1 mol of Al2S3, 0.599 mol of H2O is required

But we have 0.555 mol of H2O

so, H2O is limiting reagent

we will use H2O in further calculation

According to balanced equation

mol of H2S formed = (3/6)* moles of H2O

= (3/6)*0.555

= 0.278 mol

mass of H2S = number of mol * molar mass

= 0.278*34.08

= 9.458 g

Answer: d