What is the maximum amount of p4 that can be produced from 1.0 kg of phosphorite?

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.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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