Answer to Question #87333 in General Chemistry for Moses

Question

1) phosphorus pentachloride is produced by the reaction between phosphine and chlorine gas. What mass of chlorine gas would be required to produce 0.015kg of phosphorus pentachloride
2) if 10g of phosphine is made to react with 15g of chlorine gas. Calculate the percentage yield if 5.5g of phosphorus pentachloride was experimentally produced.

Answer 1

Task:

1) Phosphorus pentachloride is produced by the reaction between phosphine and chlorine gas. What mass of chlorine gas would be required to produce 0.015kg of phosphorus pentachloride.

2) If 10g of phosphine is made to react with 15g of chlorine gas. Calculate the percentage yield if 5.5g of phosphorus pentachloride was experimentally produced.

Solution (1):

m(PCl₅) = 0.015 kg = 15 g.

M(PCl₅) = Ar(P) + 5*Ar(Cl) = 31 + 5*35.5 = 208.5 (g/mol).

M(Cl₂) = 2*Ar(Cl) = 2*35.5 = 71 (g/mol).

The reaction between phosphine and chlorine gas:

PH₃ + 4Cl₂ = PCl₅ + 3HCl

According to the reaction equation:

n(Cl₂) / 4 = n(PCl₅)

Then,

m(Cl₂) / M(Cl₂) = 4 * m(PCl₅) / M(PCl₅)

m(Cl₂) = 4 * m(PCl₅) * M(Cl₂) / M(PCl₅)

m(Cl₂) = 4*15*71 / 208.5 = 20.43 g

Answer (1): 20.43 g of chlorine.

Solution (2):

m(PH₃) = 10 g.

M(PH₃) = Ar(P) + 3*Ar(H) = 31 + 3*1 = 34 (g/mol).

n(PH₃) = m(PH₃) / M(PH₃) = 10 / 34 = 0.2941 mol.

m(Cl₂) = 15 g.

M(Cl₂) = 2*Ar(Cl) = 2*35.5 = 71 (g/mol).

n(Cl₂) = m(Cl₂) / M(Cl₂) = 15 / 71 = 0.2113 mol.

M(PCl₅) = Ar(P) + 5*Ar(Cl) = 31 + 5*35.5 = 208.5 (g/mol).

The reaction between phosphine and chlorine gas:

PH₃ + 4Cl₂ = PCl₅ + 3HCl

We can see that 4 mol of chlorine is needed for 1 mole of phosphine. Ratio = n(PH₃) / n(Cl₂) = 1 mol / 4 mol = 0.25.

But we have the following ratio: n(PH₃) / n(Cl₂) = 0.2941 mol / 0.2113 mol = 1.39.

1.39 > 0.25

Consequently, PH₃ is in excess. Cl₂ is the limiting reactant.

n(Cl₂) / 4 = n(PCl₅)

Then,

m(Cl₂) / M(Cl₂) = 4 * m(PCl₅) / M(PCl₅)

m(PCl₅) = [M(PCl₅) * m(Cl₂)] / [4 * M(Cl₂)]

m(PCl₅) = [208.5*15] / [4*71] = 3122.25 / 284 = 10.99 g = 11 g

m_teor(PCl₅) = 11 g

w(PCl₅) = m_pract(PCl₅) / m_teor(PCl₅) = 5.5 / 11 = 0.5 or 50%.

Answer (2): the percentage yield = 50%.