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10 November, 18:19

If a particular ore contains 58.5% calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

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  1. 10 November, 18:28
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    Calcium Phospate formula: Ca3 (PO4) 2

    Atomic weight:

    Ca = 40; P = 31; O = 16

    Ca3 = 40 * 3 = 120

    P = 31

    O4 = 16 * 4 = 64

    (PO4) 2 = (31 + 64) * 2 = 95 * 2 = 190

    Ca3 (PO4) 2 = 120 + 190 = 310 g/mol

    31/310 = 10% P in the calcium phospate

    1.00kg * 1000 g/kg = 1000 g of phosphorus.

    1000 g / 10% = 10,000 g of calcium phosphate

    10,000 g / 58.5% = 17,094 grams of ore.

    The minimum mass of the ore should be 17,094 grams or 17.094 kg.
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