100 g of propane is completely reacted with 1000 g of O2. The mole fraction of carbon dioxide in the resulting mixture is (x×10−2). The value of x is

C3H8+5O2→3CO2+4H2OC3H8=10044=2.27,nO2=100032=31.2531.25−11.35=19.9X of CO2=6.8119.9+6.81+9.08=0.19=19×10−2, x=19

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100 g of propane is completely reacted with 1000 g of O₂. The mole fraction of carbon dioxide in the resulting mixture is $(x \times 10^{- 2})$ . The value of x is

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answer is 19.

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Detailed Solution

$\begin{array}{l} C_{3} H_{8} + 5 O_{2} \to 3 {CO}_{2} + 4 H_{2} O_{C_{3} H_{8}} = \frac{100}{44} = 2 .27, n_{O_{2}} = \frac{1000}{32} = 31 .25 \\ 31 .25 - 11 .35 = 19 .9 \\ X of {CO}_{2} = \frac{6 .81}{19 .9 + 6 .81 + 9 .08} = 0 .19 = 19 \times 10^{- 2}, x = 19 \end{array}$