Question

What is the maximum mass of S8 that can be produced by combining 77.0 g of each reactant?

8SO2+16H2S⟶3S8+16H2O

mass of S8

Answer 1

The given reaction is

8SO₂ + 16H₂S  $\rightarrow$ 3S₈ + 16H₂O

Lets calculate the moles of reactants

Molar mass of SO₂ = 64.06 g/mol

Molar mass of H₂S = 34.08 g/mol

using formula Moles = mass / molar mass

Moles of SO₂ = 77.0 g / 64.06 g/mol = 1.2020 mol SO₂

Moles of H₂S = 77.0 g/ 34.08 g/mol = 2.2594 mol H₂S

Now lets see moles of H₂S reacts with moles of SO₂

then ,  2.2594 mol H₂S would react completely with 2.2594 mol ( 3 mol S₈ / 16 mol H₂S ) = 0.42 mol SO₂

but as per above calculation there is more SO₂ present

So SO₂ is in excess and H₂S is limiting reactant

So

maximum mass of S₈ = 2.2594 mol H₂S $\times$ (3 mol S₈ / 16 mol H₂S) $\times$ molar mass of S₈

= 2.2594 mol H₂S $\times$ (3 mol S₈ / 16 mol H₂S) $\times$ 256.52 g/mol

= 108.67 g S₈

So our answer is 108.67 g S₈