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15 November, 22:17

If a particular ore contains 55.4 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

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  1. 16 November, 01:40
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    The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3 (PO4) 2 is 55.4%x=1000g so x=1000/0.554 = 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4 = Ca=40.078 x 3 = 120.23 and (PO4) 2 = (30.974+64) 2=189.95 (NB oxygen is 16 mass x 4 = 64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102. = 0.19 or 19% P. So of the 1.805 x 0.19 = 0.34kg of phosphorus.
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