Calculate the mass of phosphorus that can be obtained by reducing phosphorite with a mass of 193.75

| August 1, 2021 | education

Calculate the mass of phosphorus that can be obtained by reducing phosphorite with a mass of 193.75, containing 80% pure calcium phosphate, with coal. The yield of the product from the theoretically possible should be taken equal to 90%.

First, let’s create an equation:
Ca3 (PO4) 2 + 5C + 3SIO2 = 3CaSIO3 + 2P + 5CO
1) m (Ca3 (PO4) 2) = 193.75 * 80% / 100% = 155 (g) -mass of calcium phosphate in phosphate
2) n (Ca3 (PO4) 2) = m / M = 155 (g) / 310 (g / mol) = 0.5 (mol) is the amount of calcium phosphate substance
3) n (P) = 2n (Ca3 (PO4) 2) = 2 * 0.5 = 1 (mol) is the amount of phosphorus formed in this reaction. By reaction, 2 times more phosphorus is formed
4) m (P) theory. = N * M (P) = 1 (mol) * 31 (g / mol) = 31 (g) -theoretical mass of phosphorus
5) m (P) practical = m (P) theory * W / 100% = 31 (g) * 90% / 100% = 31 * 0.9 = 27.9 (g)
Answer: 27.9 (g).



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