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17 October, 22:46

If 39.0 g of C6H6 reacts with excess chlorine and produces 30.0 g of C6H5Cl in the reaction C6H6 + Cl2 → C6H5Cl + HCl, what is the percent yield of C6H5Cl?

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  1. 18 October, 01:20
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    53.4 % is the percent yield

    Explanation:

    This is the reaction:

    C₆H₆ + Cl₂ → C₆H₅Cl + HCl

    First of all we need to know the moles of benzene we used

    39 g. 1 mol / 78 g = 0.5 moles

    Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride

    0.5 moles of chloride were produced by 0.5 moles of benzene

    We must calculate the mass of chloride we produced

    0.5 mol. 112.45 g / 1 mol = 56.2g

    Let's calculate the percent yield

    (Yield produced / Theoretical yield). 100

    (30 g / 56.2 g). 100 = 53.4 %
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