JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 16)
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]
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]
Answer
19
Explanation
C3H8(g) + 5O2(g) $$\to$$ 3CO2(g) + 4H2O(l)
mole fraction of CO2 in the final reaction mixture (heterogenous)
$${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$$
= 0.1902 = 19.02 $$\times$$ 10$$-$$2
$$\Rightarrow$$ 19
| t = 0 | 2.27 mole | 31.25 mol | ||
|---|---|---|---|---|
| t = $$\infty $$ | 0 | 19.9 mol | 6.81 mol | 9.08 mol |
mole fraction of CO2 in the final reaction mixture (heterogenous)
$${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$$
= 0.1902 = 19.02 $$\times$$ 10$$-$$2
$$\Rightarrow$$ 19
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