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3 years ago

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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)?

2 answers:

Snowcat [4.5K]3 years ago

4 0

Answer is: <span>yield of a reaction is 56,4%.
</span>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%.

myrzilka [38]3 years ago

3 0

<u>Answer:</u> The percent yield of the reaction is 56.92 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of phosphorus trichloride = 200 g

Molar mass of phosphorus trichloride = 137.33 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphorus trichloride}=\frac{200g}{137.33g/mol}=1.46mol$

The chemical reaction of phosphorus trichloride with water follows the equation:

$PCl_3+3H_2O\rightarrow H_3PO_3+3HCl$

As, water is present in excess. It is considered as excess reagent.

Phosphorus trichloride is considered as a limiting reagent because it limits the formation of product.

By stoichiometry of the reaction:

1 mole of phosphorus trichloride produces 3 moles of hydrochloric acid.

So, 1.46 moles of phosphorus trichloride will produce = $\frac{3}{1}\times 1.46=4.38mol$ of hydrochloric acid.

Now, calculating the mass of hydrochloric acid from equation 1, we get:

Molar mass of hydrochloric acid = 36.5 g/mol

Moles of hydrochloric acid = 4.38 moles

Putting values in equation 1, we get:

$4.38mol=\frac{\text{Mass of hydrochloric acid}}{36.5g/mol}\\\\\text{Mass of hydrochloric acid}=159.87g$

To calculate the percentage yield of hydrochloric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrochloric acid = 91.0 g

Theoretical yield of hydrochloric acid = 159.87 g

Putting values in above equation, we get:

$\%\text{ yield of hydrochloric acid}=\frac{91.0g}{159.87g}\times 100\\\\\% \text{yield of hydrochloric acid}=56.92\%$

Hence, the percent yield of the reaction is 56.92 %.

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