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2 November, 19:23

Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?

Molecular mass of BaSO4 233.39 g/mol

Molecular mass of Ba = 137.327 g/mol

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  1. 2 November, 19:44
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    31.37%

    Explanation:

    For this case, you should consider the following reaction:

    Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

    For which you obtain the precipitate of BaSO₄

    In order to obtain the mas of Barium on the precipitate, you may use the following formula:

    gBa = M₍BaSO₄₎x (M₍Ba₎/M₍BaSO₄₎)

    Where:

    gBa = mass of Barium

    M₍BaSO₄ ₎ = mass of BaSO₄ from the precipitate

    M₍Ba₎ = mass of Barium from the original sample

    M₍BaSO₄₎ = mass of BaSO₄ from the precipitate

    gBa = (0.5331) x (137.327/233.39) = 0.3136 g

    Then we ontain the percentage of Barium multiplying by 100:

    % Ba on the original sample = 31.36%
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