Question

Phosphorus is commercially prepared by heating a mixture of calcium phosphate, sand, and coke in an electric furnace. The process involves two reactions. 2 Ca3(PO4)2(s) 6 SiO2(s) 6 CaSiO3(l) P4O10(g) P4O10(g) 10 C(s) P4(g) 10 CO(g) The P4O10 produced in the first reaction reacts with an excess of coke (C) in the second reaction. Determine the theoretical yield of P4 if 293.5 g Ca3(PO4)2 and 378.5 g SiO2 are heated. (No Response) g If the actual yield of P4 is 44.9 g, determine the percent yield of P4.

Answer 1

76.6% is percent yield of P₄

Step-by-step explanation:

Percent yield is defined as 100 times the ratio of actual yield and theoretical yield. To solve this quesiton we need to find the theoretical yield of the reaction. Using:

2Ca₃(PO₄)₂(s) + 6SiO₂(s) → 6CaSiO₃(l) + P₄O₁₀(g)

P₄O₁₀(g) + 10C(s) → P₄(g) + 10CO(g)

We need to find the moles of Ca₃(PO₄)₂ and SiO₂ to find limiting reactant. With limiting reactant we can find moles of P₄O₁₀ = Moles of P₄. We must convert the moles of P₄ to mass using Molar mass (P₄ = 123.895g/mol):

Moles Ca₃(PO₄)₂ -Molar mass: 310.1767g/mol-

293.5g * (1mol / 310.1767g) = 0.9462moles

Moles SiO₂ -Molar mass: 60.08g/mol-:

378.5g * (1mol / 60.08g) = 6.30 moles

For a complete reaction of 6.30 moles of SiO₂ there are required:

6.30 moles SiO₂ * (2 moles Ca₃(PO₄)₂ / 6 moles SiO₂) =

2.10 moles Ca₃(PO₄)₂. As there are just 0.9462 moles, Ca₃(PO₄)₂ is limiting reactant

Moles P₄O₁₀ = Moles P₄:

0.9462moles Ca₃(PO₄)₂ * (1mol P₄O₁₀ / 2 mol Ca₃(PO₄)₂) = 0.4731 moles P₄O₁₀ = Moles P₄.

The mass is:

0.4731 moles P₄ * (123.895g / 1mol) = 58.6g = Theoretical yield.

Percent yield is:

44.9g / 58.6g * 100 =

76.6% is percent yield of P₄