Find the limiting reagent and calculate the theoretical andpercent yeild for the

Question

Find the limiting reagent and calculate the theoretical andpercent yeild for the product of the following reaction.                   HC2H3O2    +    C5H12O   yeilds    C7H14O2   +   H2O    Vol.             40mL                  30mL                     5mL                MW:         60.05g/mol           88.15                    130.18 Density:      1.053g/mL         0.813                        0.879     Find the limiting reagent and calculate the theoretical andpercent yeild for the product of the following reaction.                   HC2H3O2    +    C5H12O   yeilds    C7H14O2   +   H2O    Vol.             40mL                  30mL                     5mL                MW:         60.05g/mol           88.15                    130.18 Density:      1.053g/mL         0.813                        0.879    

Explanation / Answer

1. Find limiting reagent by calculating the moles (n) of eachreagent used:     n(CH3COOH) = 1.053 g/mL * 40 mL / 60.05 g/mol = 0.7014mol     n(C5H11OH) = 0.813 g/mL *30 mL / 88.15 g/mol = 0.2767 mol     Since there are less moles ofC5H11OH (pentanol), this is the limitingreagent. 2. Calcualte theoretical yield ofC7H14O2:     From the chemical equation, which is balanced,everything reacts in a 1:1 ratio. This means:     n(C7H14O2) =n(C5H11OH) = 0.2767 mol     mass(C7H14O2) =0.2767 mol * 130.18 g/mol = 36.02g     vol(C7H14O2) =36.02 g / 0.879 g/mL = 41 mL 3. The actual yield is given as 5 mL. 4. Calculate % yield:     % Yield = actual yield / theoretical yield *100%                 = 5 / 41 * 100 = 12.2%

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