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Limiting Reagent or Reactant

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Question

In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, $C H_{3} C H O,$ under pressure at $60^{\circ} C$ in the presence of a suitable catalyst.
$2 C H_{3} C H O + O_{2} \to 2 C H_{3} C O O H$
In a laboratory test of this reaction, 20 g $C H_{3} C H O$ and 10 g $O_{2}$ were put in a reaction vessel. If the actual yield is 23.8 g, calculate the percentage yield.

A

88.15 %

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B

75.26 %

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C

65.28 %

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D

None of these

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Solution

The correct option is A 88.15 %
We know,
Molar mass of $C H_{3} C H O = 44$ g/mol

Therefore, 20 g $C H_{3} C H O = \frac{20}{44}$ mol $C H_{3} C H O = 0.45$ mol $C H_{3} C H O$

10 g $O_{2} = \frac{10}{32} m o l O_{2} = 0.31$ mol

The given balanced chemical reaction is:
$2 C H_{3} C H O + O_{2} \to 2 C H_{3} C O O H$

So, 2 moles of $C H_{3} C H O$ react with 1 mole of $O_{2}$

$∴$ 0.45 moles $C H_{3} C H O$ will react with 0.225 moles of $O_{2}$

So, moles of $C H_{3} C O O H$ produced=0.45 mol

Molar mass of $C H_{3} C O O H = 60$ g/mol

Hence, 0.45 moles $C H_{3} C O O H = 60 \times 0.45$ g = 27 g

Therefore, percentage yield = $\frac{23.8}{27} \times 100 = 88.15$ %

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