Question

What is the maximum mass of S8 that can be produced by combining 85.0 g of each reactant?

8SO2 + 16H2S —> 3S8 + 16H2O

Answer 1

Mass of SO₂ = 85 g

Molar mass of SO₂ = 64.07 g/mol

moles of SO₂ = mass/molar mass = 85g/(64.07 g/mol) = 1.33 mol

Mass of H₂S = 85 g

Molar mass of H₂S = 34.1 g/mol

Moles of H₂S = 2.49mol

From reaction, 8 mol of SO₂ reacts with 16 mol of H₂S

Thus, 1.33 mol of SO₂ reacts with 2.66 mol of H₂S

but we have 2.49mol of H₂S

So, H₂S is a limiting reagent.

16 mol of H₂S produces 3 mol of S₈

Hence 2.49mol of H₂S produces 3*2.49/16 = 0.467 mol of S₈

Molar mass of S₈ = 256g/mol

Mass of S₈ = moles*molar mass = 0.467 mol*256g/mol = 119.65 g