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kakasveta [241]
3 years ago
15

The percentage yield for the reaction

Chemistry
1 answer:
makkiz [27]3 years ago
5 0

Answer:

92.87 g.

Explanation:

∵ The percentage yield = (actual yield/theoretical yield)*100.

  • We need to calculate the theoretical yield:

From the balanced reaction:

<em>PCl₃ + Cl₂ → PCl₅,</em>

It is clear that 1 mol of PCl₃ reacts with 1 mol of Cl₂ to produce 1 mol of PCl₅.

  • We need to calculate the no. of moles of 73.7 g PCl₃:

n = mass/molar mass = (73.7 g)/(137.33 g/mol) = 0.536 mol.

<u><em>Using cross multiplication:</em></u>

1 mol of PCl₃ produce  → 1 mol of PCl₅, from stichiometry.

∴ 0.536 mol of PCl₃ produce  → 0.536 mol of PCl₅.

∴ The mass of PCl₅ (theoretical yield) = (no. of moles) * (molar mass) = (0.536 mol)*(208.24 g/mol) = 111.62 g.

<em>∵ The percentage yield = (actual yield/theoretical yield)*100.</em>

The percentage yield = 83.2%, theoretical yield = 111.62 g.

∴ The actual yield of PCl₅ = (The percentage yield)(theoretical yield)/100 = (83.2%)(111.62 g)/100 = 92.87 g.

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Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

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Mass of 0.08636 moles of ethyl butyrate =

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Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

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

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

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

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

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C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

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

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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