If you started with 50.0 grams of h2s and 60.0 grams of o2, how many grams of s8 would be produced, assuming 98 % yield?

1) Balanced chemical equation

8H2S + 4O2 - - - > S8 + 8H2O

2) Molar ratios

8 mol H2S : 4 mol O2 : 1 mol S8

3) Convert 50.0 grams of H2S into moles

moles = mass in grams / molar mass

molar mass H2S = 2*1g/mol + 32.1 g/mol = 34.1 g/mol

moles = 50.0 g / 34.1 g/mol = 1.466 moles H2S

4) Convert 60.0 g of O2 into moles

molar mass O2 = 2*16g/mol = 32.0g/mol

moles = 60.0g / 32.0 g = 1.875 moles O2

5) State limiting reagent

Theoretical ratio: 8 mol H2S / 4 mol O2 = 8/4 = 2/1 = 2

Actual ratio 1.466 mol H2S / 1.875 mol O2 < 2

=> H2S is the limiting reagent (it is consumed completely whilce some O2 remains unreacted).

6) Determine the moles of S8 produced from 1.466 moles H2S

1 mol S8 / 8 mol H2S * 1.466 mol H2S = 0.18325 mol S8

7) Convert 0.18325 mol S8 to mass

mass = number of moles * molar mass

molar mass S8 = 8 mol * 32.1g/mol = 256.8 g/mol

mass = 0.18325 mol * 256.8 g/mol = 47.1 grams

Answer: 47.1 g