Question

Consider the reaction below

Al₂S₃(s) + H₂O(l) → Al(OH)₃(s) + H₂S(g)

If 15.0g of aluminum sulfide and 10.0g of water are allowed to react as above, and assuming a complete reaction
a. by calculation, find out which is the limiting reagent.
b. calculate the maximum mass of H₂S which can be formed from these reagents.
c. calculate the mass of excess reagent remaining after the reaction is complete.

Answer 1

a)

Molar mass of Al2S3,

MM = 2*MM(Al) + 3*MM(S)

= 2*26.98 + 3*32.07

= 150.17 g/mol

mass(Al2S3)= 15.0 g

use:

number of mol of Al2S3,

n = mass of Al2S3/molar mass of Al2S3

=(15 g)/(1.502*10^2 g/mol)

= 9.989*10^-2 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 10.0 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(10 g)/(18.02 g/mol)

= 0.5551 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 9.989*10^-2 mol of Al2S3, 0.5993 mol of H2O is required

But we have 0.5551 mol of H2O

so, H2O is limiting reagent

Answer: H2O

b)

Molar mass of H2S,

MM = 2*MM(H) + 1*MM(S)

= 2*1.008 + 1*32.07

= 34.086 g/mol

According to balanced equation

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

= (3/6)*0.5551

= 0.2775 mol

use:

mass of H2S = number of mol * molar mass

= 0.2775*34.09

= 9.46 g

Answer: 9.46 g

C)

According to balanced equation

mol of Al2S3 reacted = (1/6)* moles of H2O

= (1/6)*0.5551

= 9.251*10^-2 mol

mol of Al2S3 remaining = mol initially present - mol reacted

mol of Al2S3 remaining = 9.989*10^-2 - 9.251*10^-2

mol of Al2S3 remaining = 7.376*10^-3 mol

Molar mass of Al2S3,

MM = 2*MM(Al) + 3*MM(S)

= 2*26.98 + 3*32.07

= 150.17 g/mol

use:

mass of Al2S3,

m = number of mol * molar mass

= 7.376*10^-3 mol * 1.502*10^2 g/mol

= 1.108 g

Answer: 1.11 g