Question

In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde at some pressure. The corresponding reaction is: $2C{H}_{3}CHO+O_2\to 2C{H}_{3}COOH$.
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.

• A. 0.67 L
• B. 1.34 L
• C. 0.74 L
• D. 1.48 L

Answer 1

The correct option is A 0.67 L

The balanced chemical reaction is:
$2C{H}_{3}CHO+O_2\to 2C{H}_{3}COOH$

Molar mass of $C{H}_{3}COOH=60$ g/mol
$\therefore 3.60gC{H}_{3}COOH=\frac{3.60}{60}mol=0.06mol$

$O_2$ is the limiting reagent here.
From the balanced chemical equation,
2 moles of $C{H}_{3}COOH$ is produced from 1 mole of $O_2$.
Thus 0.06 moles of $C{H}_{3}COOH$ 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 $O_2$ = $22.4\times 0.03=0.672L$