If a particular ore contains 56.3 % % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg k g of phosphoru

Question

If a particular ore contains 56.3 % % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg k g of phosphorus?

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Mộc Miên 1 month 2021-08-05T11:57:28+00:00 1 Answers 1 views 0

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    2021-08-05T11:59:13+00:00

    Answer:

    34.44 kg

    Explanation:

    First we convert 1.00 kg of phosphorus (P) into moles, using its molar mass:

    • 1.00 kg ÷ 32 kg/kmol = 0.03125 kmol P

    Then we convert 0.03125 kmoles of P into kmoles of Ca₃(PO₄)₂:

    • 0.03125 kmol P * \frac{2kmolCa_3(PO_4)_2}{2kmolP} = 0.0625 kmol Ca₃(PO₄)₂

    Now we calculate the mass of 0.0625 kmoles of Ca₃(PO₄)₂:

    • 0.0625 kmol Ca₃(PO₄)₂ * 310.18 kg/kmol = 19.39 kg

    Finally we calculate the required mass of the ore, using the definition of content percentage:

    • % content = Mass of calcium phosphate / mass of ore * 100 %
    • 56.3 % = 19.39 kg / mass of ore * 100%
    • Mass of Ore = 34.44 kg

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