14. Methanol, CH_(2)OH can be produced by the following reaction .Assume H_(2) is the limiting reactant and 4 mol of H_(2) are used in the reaction. If 1 mol of CH_(3)OH is produced by the reaction, what is the percent yield? 2H_(2)+COarrow CH_(3)OH Actual yield __ PercentYield= Theoretical yield __

To calculate the percent yield, we need to determine the theoretical yield and the actual yield.<br /><br />Given information:<br />- The balanced chemical equation is: $2H_{2} + CO \rightarrow CH_{3}OH$<br />- 4 mol of $H_{2}$ are used in the reaction.<br />- 1 mol of $CH_{3}OH$ is produced by the reaction.<br /><br />Step 1: Calculate the theoretical yield.<br />The theoretical yield is the maximum amount of product that can be formed from the given amount of reactants, assuming 100% efficiency.<br /><br />From the balanced equation, we can see that 2 mol of $H_{2}$ are required to produce 1 mol of $CH_{3}OH$.<br /><br />Theoretical yield = (1 mol $CH_{3}OH$) × (2 mol $H_{2}$ / 1 mol $CH_{3}OH$) = 2 mol $H_{2}$<br /><br />Step 2: Calculate the actual yield.<br />The actual yield is the amount of product that is actually formed in the reaction.<br /><br />Given that 1 mol of $CH_{3}OH$ is produced by the reaction, the actual yield is 1 mol.<br /><br />Step 3: Calculate the percent yield.<br />Percent yield = (Actual yield / Theoretical yield) × 100%<br />Percent yield = (1 mol / 2 mol) × 100% = 50%<br /><br />Therefore, the percent yield of the reaction is 50%.