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8 February, 12:32

A reaction was performed in which 1.400 g of camphor was reduced by an excess of sodium borohydride to make 1.061 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

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  1. 8 February, 12:43
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    The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

    Explanation:

    Step 1: Data given

    Camphor = C10H16O

    sodium borohydride = NaBH4

    Isoborneol = C10H18O

    Mass of camphor = 1.400 grams

    Mass of sodium borohydride = 1.061 grams

    Molar mass of camphor = 152.23 g/mol

    Molar mass of sodium borohydride = 37.83 g/mol

    Molar mass of isoborneol = 154.25 g/mol

    sodium borohydride is in excess, this means camphor is the limiting reactant

    Step 2: The balanced equation

    2C10H16O + NaBH4 → 2C10H18O + NaB

    Step 3: Calculate moles of camphor

    Moles camphor = mass camphor / molar mass camphor

    Moles camphor = 1.400 grams / 152.23 g/mol

    Moles camphor = 0.00920 moles

    Step 4: Calculate moles of isoborneol

    For 2 moles camphor we need 1 mol of NaBH4 to produce 2 moles of isoborneol

    For 0.00920 moles camphor we'll have 0.00920 moles isoborneol

    Step 5: Calculate mass isoborneol

    Mass isoborneol = moles isoborneol * molar mass isoborneol

    Mass isoborneol = 0.00920 * 154.25 g/mol

    Mass isoborneol = 1.4191 grams ( = theoretical yield)

    % yield = (actual yield / theoretical yield) * 100%

    % yield = (1.061 / 1.4191) * 100 %

    % yield = 74.77 %

    The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %
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