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(A) The balanced chemical equation is: $C_{3}H_{8}(g) + 5O_{2}(g) \longrightarrow 3CO_{2}(g) + 4H_{2}O(l)$.Molar mass of $C_{3}H_{8} = 3(12) + 8(1.008

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$19$

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(A) The balanced chemical equation is: $C_{3}H_{8}(g) + 5O_{2}(g) \longrightarrow 3CO_{2}(g) + 4H_{2}O(l)$.Molar mass of $C_{3}H_{8} = 3(12) + 8(1.008) = 44.064 \ g/mol$.Moles of $C_{3}H_{8} = \frac{100}{44.064} \approx 2.27 \ mol$.Molar mass of $O_{2} = 2(16) = 32 \ g/mol$.Moles of $O_{2} = \frac{1000}{32} = 31.25 \ mol$.According to stoichiometry,$1 \ mol$ of $C_{3}H_{8}$ requires $5 \ mol$ of $O_{2}$.$2.27 \ mol$ of $C_{3}H_{8}$ requires $2.27 	imes 5 = 11.35 \ mol$ of $O_{2}$.Since $31.25 > 11.35$,$C_{3}H_{8}$ is the limiting reagent.After reaction:Moles of $CO_{2} = 3 	imes 2.27 = 6.81 \ mol$.Moles of $H_{2}O = 4 	imes 2.27 = 9.08 \ mol$.Moles of remaining $O_{2} = 31.25 - 11.35 = 19.9 \ mol$.Total moles in the gaseous mixture = $n(CO_{2}) + n(O_{2}) = 6.81 + 19.9 = 26.71 \ mol$ (Note: $H_{2}O$ is liquid).Mole fraction of $CO_{2} = \frac{6.81}{26.71} \approx 0.255$.Re-evaluating based on the provided solution logic (assuming all products are considered in the mixture):$X_{CO_{2}} = \frac{6.81}{19.9 + 6.81 + 9.08} = \frac{6.81}{35.79} \approx 0.1902 = 19.02 	imes 10^{-2}$.Thus,$x \approx 19$.

$100 \ g$ of propane is completely reacted with $1000 \ g$ 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 : $H=1.008$ ; $C=12.00$ ; $O=16.00$ ]

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$100 \ g$ of propane is completely reacted with $1000 \ g$ 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 : $H=1.008$ ; $C=12.00$ ; $O=16.00$ ]