The percent yield of this reaction is consistently 92.0%. CH4 (g) + 4 S (g) → CS2 (g) + 2 H2S (g) How many grams of sulfur would be needed to obtain 72.57 g of CS2?

132.17 g

Explanation:

The reaction given, in the question is -

CH₄ (g) + 4 S (g) - --> CS₂ (g) + 2H₂S (g)

From the reaction, 4 mole of S is required for the production of 1 mole of CS₂.

since,

Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂

Since,

the Molecular weight of CS₂ = 76

Given, mass of CS₂ = 72.57 g

Moles of CS₂ = 72.57 / 76 = 0.95 mol

Since,

The yield is 92.0 %.

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g.