In a process for producing acetic acid- oxygen gas is bubbled into acetaldehyde at some pressure- The corresponding reaction is- 2CH3CHO-O2- 2CH3COOH-In a labortory test of this reaction where oxygen supply is limited- if 3-60 g of acetic acid is produced- calculate the volume of oxygen gas at STP-

Name: JEE - Grade 11 - Chemistry - Chemical bonding - Session 01 - W01
Uploaded: 2022-07-04T10:36:42+05:30
Description: The balanced chemical reaction is: 2CH_3CHO+O_2→ 2CH_3COOH Molar mass of CH_3COOH = 60 g/mol ∴ 3.60 g CH_3COOH = 3.60/60 mol = 0.06 mol O_2 is the limiting ...

The balanced chemical reaction is: 2CH_3CHO+O_2→ 2CH_3COOH Molar mass of CH_3COOH = 60 g/mol ∴ 3.60 g CH_3COOH = 3.60/60 mol = 0.06 mol O_2 is the limiting ...

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Standard XII

Chemistry

Mole Concept

In a process ...

Question

In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde at some pressure. The corresponding reaction is: $2 C H_{3} C H O + O_{2} \to 2 C H_{3} C O O H$ .
In a labortory test of this reaction where oxygen supply is limited, if 3.60 g of acetic acid is produced, calculate the volume of oxygen gas at STP.

0.67 L

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1.34 L

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0.74 L

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1.48 L

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Solution

The correct option is A 0.67 L
The balanced chemical reaction is:
$2 C H_{3} C H O + O_{2} \to 2 C H_{3} C O O H$

Molar mass of $C H_{3} C O O H = 60$ g/mol
$∴ 3.60 g C H_{3} C O O H = \frac{3.60}{60} m o l = 0.06 m o l$

$O_{2}$ is the limiting reagent here.
From the balanced chemical equation,
2 moles of $C H_{3} C O O H$ is produced from 1 mole of $O_{2}$ .
Thus 0.06 moles of $C H_{3} C O O H$ will be produced from 0.03 moles of $O_{2}$ .

Again, at STP,
the volume of 1 mole of $O_{2}$ = 22.4 L
Thus the volume of 0.03 moles of $O_{2}$ = $22.4 \times 0.03 = 0.672 L$

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