Question

If 225 g of carbon reacts with excess sulfur dioxide to produce 195 g of carbon disulfide, what is the percent yield for the reaction? SC+2 SO2 → CS2 +4 CO (mwt: CS2 = 76.139 g/mol, co = 28.01 g/mol, C = 12 g/mol, SO2 = 64.066 g/mol) 78.9% a. Ob 22.5% Oc 19.5% Od. 68.4% 15.7% Oe.

Answer 1

68.3% (option d)

Step-by-step explanation:

Given, 5C+ 2SO2 → CS2 + 4CO
5 moles of C reacts with 2 moles of SO2 to produce 1 mole of CS2 and 4 moles of CO.

We have 225 grams of carbon (12 g/mol) ⇒ 225/12 moles of carbon

Now, we calculate the theoretical yield, with carbon as the limiting reagent:
5 moles of C reacts to produce 1 mole of carbon disulphide
225/12 moles of C produces 225/(12*5) = 15/4 moles of Carbon Disulphide
(15/4) * 76.139 = 285.52125 grams

But the actual yield is just 195 grams

We now find the yield % = (195/285.52125) * 100
= 68.3%