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 .

the answer is barium and copper bromide