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15 July, 18:28

A 5.18-g sample of a mixture of MgBr2 (s) and sucrose (C12H22O11) is dissolved in water and all the bromide is precipitated as AgBr (s). If the AgBr (s) precipitate has a mass of 7.65 g, what is the mass percent of MgBr2 (s) in the mixture?

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  1. 15 July, 18:40
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    Answer: 62.8%

    Explanation:

    step 1: Calculate the number of moles (n) for AgBr.

    n (AgBr) = mass (AgBr) / Molar mass (AgBr)

    = (7.65g) / (187.77g/mol)

    = 0.0407 mol

    step 2: Determine the number of moles of Br in AgBr.

    1 mol AgBr = 1 mol of Br

    therefore; n (Br) = 0.0407mol

    step 3: Calculate the mass of Br precipitated.

    m (Br) = n (Br) * Molar mass (Br)

    = (0.0407 mol) / (79.904g/mol)

    = 3.255g

    step 4: Calculate the mass percent of MgBr₂.

    %MgBr₂ = mass (Br) / mass (MgBr₂ + C₁₂H₂₂O₁₁) * 100

    = (3.225g / 5.18g) * 100

    = 62.8%
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