Percent Yield

% yield = actual yield / theoretical yield x 100

C6H6(l) + Cl2(g) ? C6H5Cl(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) ? CH3OH(l)

If 75.0g of carbon monoxide reacts with excess hydrogen gas to produce 68.4g of methanol, what is the percent yield of methanol?

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