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$100 \mathrm{~g}$ of propane is completely reacted with $1000 \mathrm{~g}$ of oxygen. The mole fraction of carbon dioxide in the resulting mixture is $x \times 10^{-2}$ . The value of $x$ is____________.(Nearest integer) $[Atomic\: \: weight : \mathrm{H}=1.008 ; \mathrm{C}=12.00 ; \mathrm{O}=16 . \mathrm{00}]$

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Moles of $C_{3}H_{8}= \frac{100}{44}= 2.27$ , moles of $O_{2}= \frac{1000}{32}= 31.25$

Thus, propane is the limiting reagent.

The combustion reaction is given as :

$\mathrm{C_{3}H_{8}\quad+\quad5O_{2}\quad \quad \rightarrow\quad 3CO_{2}\quad+\quad 4H_{2}O}$

$2.27$ $31.25$

$-$ $31.25-5\left ( 2.27 \right )$ $3\left ( 2.27 \right )$ $4\left ( 2.27 \right )$

$= 19.9$ $= 6.81$ $= 9.08$

$\therefore$ Mole fraction of $CO_{2}= \frac{6.81}{19.9+6.81+9.08}$

$= \frac{6.81}{35.79}$

$= 19.02\times10^{-2}$

Hence the answer is 19

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