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27 August, 08:35

If you start with 10.0 grams of lithium hydroxide and an excess of hydrogen bromide, how many grams of lithium bromide will be produced?

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  1. 27 August, 08:48
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    grams of lithium bromide produced = 36.2642309286 ≈ 36.30 g

    Explanation:

    The equation of the reaction is as follows;

    Lithium hydroxide + hydrogen bromide → Lithium Bromide + water

    LiOH + HBr → LiBr + H2O

    The equation is already balanced

    Molecular mass of LiOH = 6.941 + 15.999 + 1.00784

    Molecular mass of LiOH = 23.94784 g

    Molecular mass of LiBr = 6.941 + 79.904 = 86.845 g

    From the equation

    LiOH + HBr → LiBr + H2O

    23.94784 g of lithium hydroxide produces 86.845 g of lithium bromide

    10 g of lithium hydroxide will produce?

    cross multiply

    grams of lithium bromide produced = 10 * 86.845/23.94784

    grams of lithium bromide produced = 868.45/23.94784

    grams of lithium bromide produced = 36.2642309286 ≈ 36.30 g
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