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13 May, 12:14

Using the balanced equation below, how many grams of lead (II) sulfate would be produced from the complete reaction of 23.6 g lead (IV) oxide?

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  1. 13 May, 14:02
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    The chemical equation is:

    Pb + PbO2 + 2 H2SO4 → 2 PbSO4 + 2 H2O

    From the periodic table:

    mass of Pb = 207.2 grams

    mass of oxygen = 16 grams

    mass of sulfur = 32 grams

    Therefore:

    molar mass of PbO2 = 207.2 + 2 (16) = 239.2 grams

    molar mass of PbSO4 = 207.2 + 32 + 4 (16) = 303.2 grams

    From the balanced equation, one mole of PbO2 produces two moles of PbSO4. This means that 239.2 grams of PbO2 produces 606.4 grams of PbSO4.

    Use cross multiplication to find the amount of PbSO4 produced from 23.6 grams of PbO2 as follows:

    amount = (23.6*606.2) / (239.2) = 59.8287 grams
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