Question

What is the maximum mass of S 8 that can be produced by combining 95.0 g of each reactant?

8 SO 2 + 16 H 2 S ⟶ 3 S 8 + 16 H 2 O

Answer 1

Molar mass of SO2,
MM = 1*MM(S) + 2*MM(O)
= 1*32.07 + 2*16.0
= 64.07 g/mol


mass(SO2)= 95.0 g

use:
number of mol of SO2,
n = mass of SO2/molar mass of SO2
=(95 g)/(64.07 g/mol)
= 1.483 mol

Molar mass of H2S,
MM = 2*MM(H) + 1*MM(S)
= 2*1.008 + 1*32.07
= 34.086 g/mol


mass(H2S)= 95.0 g

use:
number of mol of H2S,
n = mass of H2S/molar mass of H2S
=(95 g)/(34.09 g/mol)
= 2.787 mol
Balanced chemical equation is:
8 SO2 + 16 H2S ---> 3 S8 + 16 H2O


8 mol of SO2 reacts with 16 mol of H2S
for 1.483 mol of SO2, 2.966 mol of H2S is required
But we have 2.787 mol of H2S

so, H2S is limiting reagent
we will use H2S in further calculation


Molar mass of S8 = 256.56 g/mol

According to balanced equation
mol of S8 formed = (3/16)* moles of H2S
= (3/16)*2.787
= 0.5226 mol


use:
mass of S8 = number of mol * molar mass
= 0.5226*2.566*10^2
= 1.341*10^2 g
Answer: 134 g