Answer and explanation;
-Phosphorous can be prepared from calcium phosphate by the following reaction: 2Ca3(PO4)2 + 6SiO2 + 10C -----> 6CaSi03 + P4 + 10CO
Phosphorite is a mineral that contains Ca3(PO4)2 plus other non phosphorous containing compounds.
-If the phosphorite sample is 75% Ca3(PO4)2 by mass, then;
mass of Ca3(PO4)2 is given by; 0.75 * (1000 g) = 750g
MM Ca3(PO4)2 = 310.19 g/mol
thus, Number of moles of Ca3(PO4)2;
= 750g / 310.19 g/mol
= 2.42 moles of Ca3(PO4)2 From the equation;
(2 mol Ca3(PO4)2 = 1 mol P4) Thus; moles of P4 will be;
=2.42 mol Ca3(PO4)2 * (1 mol P4 / 2 mol Ca3(PO4)2)
= 1.21 moles of P4 Mass of P4 will be;
=1.21 mol P4 * ( 4 x 30.974) g/mol
= 1.21 mol P4 * 123.896 g/mol
= 150 g P4. (2 sf)
Hence;
1.5 x10^(2) g P4 are produced from 1.0 kg Phosphorite mineral containing 75% Ca3(PO4)2.
