Calculate the mass of phosphorite containing 62% of calcium orthophosphate required to obtain 3.1 kg

| September 12, 2021 | education

Calculate the mass of phosphorite containing 62% of calcium orthophosphate required to obtain 3.1 kg of phosphorus, if the phosphorus yield is 92% of the theoretically possible.

1.write the reaction of obtaining phosphorus:

2Ca3 (PO4) 2 + 6SiO2 + 10C = 6CaSiO3 + P4 + 10CO;

2.Calculate the practical chemical amount of phosphorus obtained:

npract (P) = m (P4): M (P4) = 3100: 124 = 25 mol;

3. Determine the theoretical amount of phosphorus:

ntheor (P) = ntract (P): ν = 25: 0.92 = 27.17 mol;

4. find the amount and mass of calcium phosphate:

n (Ca3 (PO4) 2) = ntheor (P) * 2 = 27.17 * 2 = 54.34 mol;

m (Ca3 (PO4) 2) = n (Ca3 (PO4) 2) * M (Ca3 (PO4) 2);

M (Ca3 (PO4) 2) = 40 * 3 + 2 * 31 + 8 * 16 = 310 g / mol;

m (Ca3 (PO4) 2) = 54.34 * 310 = 16845 g = 16.845 kg;

5.Calculate the mass of phosphorite:

m (phosphorite) = m (Ca3 (PO4) 2): w (Ca3 (PO4) 2) = 16.845: 0.62 = 27.17 kg.

Answer: 27.17 kg.



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