Question

5L of an alkane requires 25L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is:
a.) Butane
b.) Isobutane
c.) Propane
d.) Ethane

Answer 1

Hint: Try to recall the standard reaction for the combustion of alkane. The combustion reaction of alkane forms the product carbon dioxide and water. Now you can easily find the alkane and get the answer to this question.

Complete step by step answer:

We can write the reaction for the combustion of alkane as,

$C_{ n }H_{ 2n+2 }\quad +\quad (\dfrac { 3n+1 }{ 2 } )O_{ 2 }\quad \rightarrow \quad nCO_{ 2 }\quad +\quad (n+1)H_{ 2 }O$

From the above reaction, we can determine that,
No. of moles of alkane = 1
No. of moles of oxygen = $(\dfrac { 3n+1 }{ 2 } )$
​Now, from the ideal gas equation, we know
PV = nRT
At the same temperature and pressure,
V ∝ n
Or we can write this as,
V(alkane)n(oxygen) = V(oxygen)n(alkane) …….(I)
Now we have given the following in the question,
The volume of alkane = 5L
The volume of oxygen = 25L
From equation(I), we have
5×$(\dfrac { 3n+1 }{ 2 } )$ = 25×1
​⇒3n+1=5×2
⇒3n=10−1
⇒n=3
Thus the alkane is $C_{ 3 }H_{ 2×3+2 }\quad =\quad C_{ 3 }H_{ 8 }$
​That means this is propane.
We can write the combustion reaction of propane like this,
$C_{ 3 }H_{ 8 }\quad +\quad 5O_{ 2 }\quad \rightarrow \quad 3CO_{ 2 }\quad +\quad 4H_{ 2 }O$
Here, 1 L of propane required 5 L of oxygen to completely burn. Hence, 5 L of propane requires 5×5 = 25 L oxygen for complete combustion.

Therefore, the correct answer to this question is option B.

Note: We should be also familiar with the general equation of combustion for any hydrocarbon. Let $C_{ x }H_{ y }$ represent our hydrocarbon. It doesn’t matter whether it is an alk-ane, -ene, or -yne. The equation for the combustion reaction will be:
$C_{ x }H_{ y }\quad +\quad (x\quad +\dfrac { y }{ 4 } )O_{ 2 }\quad \rightarrow \quad xCO_{ 2 }\quad +\quad \dfrac { y }{ 2 } H_{ 2 }O$

Here x, is the number of carbons and y, is the number of hydrogens present in the hydrocarbon.