Question

If you started with 50.0 grams of H2S and 40.0 grams of O2, how many grams of S8 would be produced, assuming 97 % yield?

Answer 1

The above is the balanced reaction between H2S and O2 forming S8

Molar mass of H2S = 2 * 1 + 32 = 34 gm/mol

Number of moles of H2S = 50/34 = 1.4705 moles

Molar mass of O2 = 32 gm/mol

Number of moles of O2 = 40/32 = 1.25 moles

8 moles of H2S reacts with 4 moles of O2. hence H2S is the limiting reagent in the reaction

8 moles of H2S will form 1 mole of S8

Moles of S8 formed = 1.4705/8 = 0.1838125 moles

Molar mass of S8 = 32 * 8 = 256 gm/mol

Theoritical Yield = Number of moles * molar mass = 0.1838125 moles * 256 gm/mol = 47.056 gms

Assuming 97% Yield

Mass of S8 Produced = 47.056 * 97/100 = 45.644 gms