Question

What is the maximum mass of S8 that can be produced by combining 95.0 g of each reactant?

8SO2+16H2S⟶3S8+16H2O

Answer 1

Number of moles of SO2 = 95.0 g / 64.044 g/mol = 1.48 mole

Number of moles of H2S = 95.0 g / 34.1 g/mol = 2.79 mole

From the balanced equation we can say that

8 mole of SO2 requires 16 mole of H2S so

1.48 mole of SO2 will require

= 1.48 mole of SO2 *(16 mole of H2S / 8 mole of SO2)

= 2.96 mole of H2S

But we have 2.79 mole of H2S which is in short so H2S is limiting reactant

From the balanced equation we can say that

16 mole of H2S produces 3 mole of S8 so

2.79 mole of H2S will produce

= 2.79 mole of H2S *(3 mole of S8 / 16 mole of H2S)

= 0.523 mole of S8

mass of 1 mole of S8 = 256 g so

the mass of 0.523 mole of S8 = 134 g

Therefore, the mass of S8 produced would be 134 g