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29 March, 17:10

Lead is found in Earth's crust in several different lead ores. Suppose a certain rock is composed of 38.0% PbS (galena), 25.0% PbCO3 (cerussite), and 17.4% PbSO4 (anglesite). The remainder of the rock is composed of substances that don't contain lead. How much of this rock (in kg) must be processed to obtain 5.8 metric tons of lead? (A metric ton is 1000 kg.)

Express your answer using two significant figures.

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  1. 29 March, 20:13
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    For each compound, i. e. galena, cerussite and anglesite calculate the quantity of Pb contained.

    For that, you need to use the molecular formula and the atomic weights of the elements.

    Galena, PbS

    Atomic weights

    Pb: 207 uma

    S: 32 uma

    Molar mass = 207 + 32 = 239 g/mol

    Proportion of Pb in the compound = 207/239

    % of Pb from galena: 38.0% * [207/239]

    Cerussite

    Pb CO3

    Atomic weights

    Pb: 207

    C: 12

    O: 16

    Molar mass = 207 + 12 + 3*16 = 267 g/mol

    % of Pb coming from the cerussite

    25.0% * [207/267]

    Anglesite, PbSO4

    Molar mass: 207 + 32 + 4*16 = 303 g/mol

    % of Pb coming from anglesite = 17.4% * [207/303]

    Total quantity of Pb from the rock (call X the quantity of rock)

    [38.0%*207/239 + 25.0%*207/267 + 17.4%*207/303]*X.

    And that must be equal to 5.8 metric tons = 5800 kg

    The result of the calculations are:

    [32.91% + 19.38% + 11.89%]X = 5800 KG

    64.18% X = 5800 KG

    0.6418 X = 5800 KG

    X = 5800 / 0.6418

    X = 9037 KG

    Which you have to express with two significan figures, i. e. 9000 kg
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