what is the percentage yield of carbon disulfide if the reaction of 32.0g of sulfur dioxide produces 12.0g of carbon disulfide?

Answer;

= 63.04 %

Explanation;

The balanced equation for the reaction;

5 C + 2 SO2 → CS2 + 4 CO

We can calculate the theoretical yield;

= (32.0 g SO2) / (64 g C/mol) x (1/2) x (76.1418 g CS2/mol)

= 19.035 g CS2 in theory

But;

Percentage yield = Actual yield/theoretical yield * 100%

= (12.0 g) / (19.035 g) * 100%

= 0.6304 * 100%

= 63.04%