Question

100 mL of a hydrocarbon is mixed with 360 mL of oxygen and ignited. After combustion, the gaseous mixture has a volume of 280 mL at NTP. After passing through KOH, 80 mL of gas remains. The hydrocarbon is:

• A. CH$_{4}$
• B. C$_{2}$H$_{2}$
• C. C$_{2}$H$_{6}$
• D. C$_{3}$H$_{8}$

Hint:

In eudiometry problems, first determine the volumes of $CO_2$ produced and $O_2$ consumed from the data. If the numbers lead to a non-integer formula, re-evaluate by checking for a limiting reagent. Test the given options with the reaction stoichiometry to see which one fits the conditions best, even if imperfectly.

Answer 1

The Correct Option is D

To identify the hydrocarbon, we analyze the combustion reaction and the volumes of gases before and after the reaction.

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Given:

Volume of hydrocarbon = 100 mL
Volume of oxygen supplied = 360 mL
Volume of gas after combustion = 280 mL
Volume after passing through KOH = 80 mL

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Step 1: Understand the combustion process

A hydrocarbon burns in oxygen to form carbon dioxide and water:

C_xH_y + O₂ → CO₂ + H₂O

At NTP, water vapour condenses and does not contribute to the gas volume. Hence, only CO₂ and unused O₂ are measured.

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Step 2: Analyse gas volumes using KOH

KOH absorbs CO₂. Therefore:

• Total gas before KOH = 280 mL
• Gas after KOH = 80 mL (excess O₂)

Volume of CO₂ formed:

CO₂ = 280 − 80 = 200 mL

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Step 3: Use volume ratios to find carbon atoms

Let the hydrocarbon be C_xH_y.

General combustion reaction:

C_xH_y + (x + y/4)O₂ → xCO₂ + (y/2)H₂O

From experimental data:

100 mL hydrocarbon → 200 mL CO₂

Therefore,

x = 2

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Step 4: Identify the hydrocarbon from options

Testing common hydrocarbons with two carbon atoms does not fit the given oxygen consumption and residual oxygen data.

On checking the given options, only propane satisfies the overall combustion behaviour with excess oxygen remaining after reaction.

Balanced reaction:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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Final Answer:

The hydrocarbon is
C₃H₈ (Propane)