Question

If a LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be rquired for combustion of $1$ kg of it will be:

• A. $2.50\times {10}^3$g
• B. $4.50\times {10}^3$g
• C. $1.80\times {10}^3$g
• D. $3.58\times {10}^3$g

Answer 1

The correct option is D $3.58\times {10}^3$g

The standard enthalpy of combustion of butane $C_4{H}_{10}$ (both n- and iso butanes are main constituents of cooking gas), representing the combustion of $1$ mole of butane in the presence of oxygen, can be represented as follows.

$\underset{1mol}{C_4{H}_{10}}\left(g\right)+\frac{13}{2}{O}_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(l\right)$

$\left(58g\right)$ of $C_4{H}_{10}$ as per the reaction rquires $\left(\frac{13}{2}\times 32=208g\right)$ of $O_2$

$\therefore O_2$ required for $1kg$ butane

$=\frac{208}{58}\times 1000g$

$=3.58\times {10}^{3}g$.