What is the percent yield (omit % symbol in the answer) if reaction of 42.3 g of carbon yields 44.7 g of carbon disulfide? 5 C(s) + 2 SO2 (9) CS2 (9) + 4CO (9)

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

In such a way, since 44.7 g of carbon disulfide (76 g/mol) are actually yielded, we need to compute the theoretical yield via the 42.3 g of carbon (12 g/mol) and the 1:5 mole ratio between them as shown below:

Thus, the percent yield, which is computed by dividing the actual yield over the theoretical one, we obtain:

Best regards.

the answer is barium and copper bromide