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What is the percent yield of a reaction in which 200. g of phosphorus trichloride reacts with excess water to form 91.0 g of hcl and aqueous phosphorous acid (h3po3)?

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Dejanras
Answer is: yield of a reaction is 56,4%.
Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.
Eduard22sly

The percentage yield of HCl obtained from the reaction is 57.1%

We'll begin by calculating the mass of PCl₃ that reacted and the mass of HCl produced from the reaction.

PCl₃ + 3H₂O → 3HCl + H₃PO₃

Molar mass of PCl₃ = 31 + (3×35.5) = 137.5 g/mol

Mass of PCl₃ from the balanced equation = 1 × 137.5 = 137.5 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Mass of HCl from the balanced equation = 3 × 36.5 = 109.5 g

SUMMARY:

From the balanced equation above,

137.5 g of PCl₃ reacted to produce 109.5 g of HCl

  • Next, we shall determine the theoretical yield of HCl.

From the balanced equation above,

137.5 g of PCl₃ reacted to produce 109.5 g of HCl.

Therefore,

200 g of PCl₃ will react to produce = k200 × 109.5) / 137.5 = 159.3 g of HCl.

  • Finally, we shall determine the percentage yield.

Actual yield of HCl = 91 g

Theoretical yield of HCl = 159.3 g

Percentage yield =?

[tex]percentage \: yield = \frac{actual}{theoretical} \times 100 \\ \\ percentage \: yield = \frac{91}{159.3} \times 100 \\ \\ [/tex]

Percentage yield of HCl = 57.1%

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