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3 July, 00:56

If a particular ore contains 55.1 % calcium phosphate, what minimum mass of the ore must be processed to

obtain 1.00 kg of phosphorus?

Express your answer with the appropriate units.

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Answers (1)
  1. 3 July, 03:16
    0
    6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.

    Explanation:

    Given

    An ore has 51% Calcium phosphate

    To find

    The minimum mass of ore to be processed to get 1.00 Kg of phosphorous

    First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃ (PO₄) ₃.

    Molar mass of Ca₃ (PO₄) ₃ is 310 g

    Molar mass of P is 31 g

    1 mole of Ca₃ (PO₄) ₃ has 3 atoms of phosphorous

    i. e 310 g of Ca₃ (PO₄) ₃ has 3 * 31g of P

    = 93 g P

    93 g P wil be present in 310 g of Ca₃ (PO₄) ₃

    1 Kg or 1000 g of P will be in (1000:93) * 310

    =3333.33 g of Ca₃ (PO₄) ₃

    but the ore has only 55.1% Ca₃ (PO₄) ₃

    i. e

    100 g of Ca₃ (PO₄) ₃ will have 55.1g Ca₃ (PO₄) ₃

    we need 3333.33g of Ca₃ (PO₄) ₃

    100 g of ore will have 55.1g Ca₃ (PO₄) ₃

    3333.33g of Ca₃ (PO₄) ₃ will be present in

    (3333.33: 55.1) * 100

    = 6049.60 g of the ore

    So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
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