The mineral phosphorite (Ca3(PO4)2) exists as phosphate rock in its impure form. Elemental phosphorous is prepared from phosphate rock by reduction by carbon in the presence of sand:
2Ca3(PO4)2 + 10C + 6 SiO2 !
P4 + 10CO + 6CaSiO3.
How many phosphorous molecules (P4) can be obtained from 6 g of phosphate rock that is 90% pure phosphorite?
Answer in units of moles
2Ca3(PO4)2 + 10C + 6 SiO2 !
P4 + 10CO + 6CaSiO3.
How many phosphorous molecules (P4) can be obtained from 6 g of phosphate rock that is 90% pure phosphorite?
Answer in units of moles
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(.90)(6g)=5.4g pure phosphorous
(5.4g)(1mol P4/123.88g P4)(6.022x10^23molecules/1mol)=
4.86x10^21 molecules P4
(5.4g)(1mol P4/123.88g P4)(6.022x10^23molecules/1mol)=
4.86x10^21 molecules P4