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27 February, 11:06

If a particular ore contains 58.9 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

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  1. 27 February, 14:37
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    1 mole of phosphorous contains 31 g / mol

    Thus, 1000 g will contain 1000/31 moles

    The RFM of calcium phosphate (Ca₃ (PO₄) ₂ = 310 G

    1 Mole of calcium phosphate contains 2 moles of P

    Therefore, the number of moles of Calcium phosphate that will contain 1000/31 moles will be;

    (1/2) * (1000/31) = 1000/62 moles

    but 1 mole of calcium phosphate = 310 g

    thus, (1000/62) moles calcium phosphate will contain;

    = (1000/62) * 310

    = 5000 g or 5 kg

    Therefore, for 5 kg of calcium phosphate to be processed we will need

    (5000 / 58.9) * 100

    = 8488.96 g or 8.489 kg of the ore

    Thus, a minimum of 8.489 kg of the ore must be processed to yield 1 kg of phosphorus
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