Answer to Question #259524 in General Chemistry for Cozy

Question

Calcium phosphate reacts with sand and carbon to form phosphorus and carbon monoxide and if the reaction is 78.50 %efficient, calculate the mass of phosphorus that maybe be produced from 2.35 kg calcium phosphate, 1.00 kg SiO2 and an excess C. Apart from carbon, what mass of the other excess reactant remained at the end of the reaction?

Answer 1

Ca₃(PO₄)₂ + 5 C + 3 SiO₂ → 2 P + 3 CaSiO₃ + 5 CO

n = m / M

M (Ca₃(PO₄)₂) = 310.18 g/mol

M (SiO₂) = 60.08 g/mol

M (P) = 30.97 g/mol

M (CaSiO₃) = 172.24 g/mol

Available amount of key reactants is:

n (Ca₃(PO₄)₂) = 2 350 / 310.18 = 7.58 mol

n (SiO₂) = 1 000 / 60.08 = 16.64 mol

According to the equation, n (P) = 2 x n (Ca₃(PO₄)₂) = 2/3 x n (SiO₂). The amount of Phosphorus to be formed from the given amount of key reactants:

n (P) = 2 x n (Ca₃(PO₄)₂) = 2 x 7.58 = 15.16 mol

n (P) = 2/3 x n (SiO₂) = 2/3 x 16.64 = 11.1 mol

Therefore, SiO₂ is the limiting reactant.

The amount of Phosphorus being formed is: n (P) = 11.1 mol.

m (P) = n x M = 11.1 x 30.97 = 343.77 g

m (P)_78.50% = 343.77 x 0.7850 = 269.86 g

n (P)_78.50% = 269.86 / 30.97 = 8.71 mol

The remaining amount of Ca₃(PO₄)₂ is:

n (Ca₃(PO₄)₂)_excess = n (Ca₃(PO₄)₂) - (n (P) / 2) = 7.58 - (8.71 / 2) = 3.22 mol

m (Ca₃(PO₄)₂)_excess = n x M = 3.22 x 310.18 = 998.78 g