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13 March, 17:28

What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?

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  1. 13 March, 20:11
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    1) Chemical reaction

    C6H6 + Br2 - - - > C6H5Br + HBr

    2) Molar raios

    1 mol C6H6 : 1 mol Br2 : 1 mol C6H5Br

    3) Convert the data to moles

    42.1 g of C6H6

    molar mass of C6H6 = 6*12g/mol + 6*1g/mol = 78 g/mol

    42.1 g / 78 g/mol = 0.54 mol

    73.0 g of Br2

    molar mass of Br2 = 2 * 79.9 g/mol = 159.8 g/mol

    73.0 g / 159.8 g/mol = 0.46 mol of Br2

    => limiting reagent is the Br2.

    4) Product

    1 mol of Br2 (limiting reagent) yields 1 mole of C6H5Br, then you will obtain 0.46 mol of C6H5Br

    5) Convert the product to grams

    molar mass of C6H5Br = 6*12g/mol + 5*1g/mol + 79.9 g/mol = 156.9 g/mol

    0.46mol*156.9g/mol = 72.2 g.

    Answer: 72.2 g
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