What volume of oxygen gas (O_2) measured at 0^ C and 1 atm, is needed to burn completely 1 ~L of propane gas (C_3 H_8) measured under the same conditions?

Correct answer: D. 5 ~L aligned & 22.4 ~L at STP C_ 3 H_ 8 (~g) + 5 ×22.4 ~L at STP 5 O_ 2 ( ~g) → 3 CO_ 2 (~g)+4 H_ 2 O(I) & 22.4 ~L C_ 3 H_ 8 at STP 5 ×22.4 ~L of O_ 2 at STP & 1 LC_ 3 H_ 8 at STP (5 × 22.4)/(22.4) of O_ 2 at STP & =5 ~L of O_ 2 at NTP & aligned

NEET 2008

Exam: NEET
Year: 2008
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

NEET 2008 · Chemistry · Some Basic Concepts of Chemistry

What volume of oxygen gas (O_2) measured at 0^ C and 1 atm, is...

What volume of oxygen gas

\left(\mathrm{O}_2\right)

measured at

0^{\circ} \mathrm{C}

and 1 atm, is needed to burn completely

1 \mathrm{~L}

of propane gas

\left(\mathrm{C}_3 \mathrm{H}_8\right)

measured under the same conditions?

Options

A $10 \mathrm{~L}$
B $7 \mathrm{~L}$
C $6 \mathrm{~L}$
D $5 \mathrm{~L}$

Correct answer

D. $5 \mathrm{~L}$

Step-by-step solution

\begin{aligned}& \underset{22.4 \mathrm{~L} \text{ at STP }}{\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})}+\underset{5 \times22.4 \mathrm{~L} \text{ at STP }}{5 \mathrm{O}_{2}( \mathrm{~g})} → 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})\\ & \because 22.4 \mathrm{~L} \mathrm{C}_{3} \mathrm{H}_{8} \text{ at STP } \equiv5 \times22.4 \mathrm{~L} \text{ of } \mathrm{O}_{2} \text{ at STP }\\ & \therefore 1 \mathrm{LC}_{3} \mathrm{H}_{8} \text{ at STP } \equiv \frac{5 \times 22.4}{22.4} \text{ of } \mathrm{O}_{2} \text{ at STP }\\ & =5 \mathrm{~L} \text{ of } \mathrm{O}_{2} \text{ at NTP }\\ & \end{aligned}

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