% yield = actual yield / theoretical yield x 100
(l) + Cl2
(g) ? C6
Cl(s) + HCl(g)
When 36.8 g of C6H6 react with excess Cl2, 38.8g of C6H5Cl(s) are produced (the actual yield). What is the percent yield of C6H5Cl(s)?
CO(g) + 2H2
(g) ? CH3
If 75.0g of carbon monoxide reacts with excess hydrogen gas to produce 68.4g of methanol, what is the percent yield of methanol?