100 g of propane is completely reacted with 1000g of oxygen-The mole fraction of carbon dioxide in the resulting mixture is x- 10-2-The value of -x- is Rounded off to the nearest integer is -Atomic weight -H-1-008-C-12-00-O-16-00-

100 g of propane = 2.27mol 1000g oxygen= 31.25 mol C_3H_8(g)+5O_2(g)→ 3CO_2(g)+4H_2O (l) From the equation 1 mol of propane reacts with 5 mol of oxygen to give ...

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Standard XII

Chemistry

Molarity

100 g of prop...

Question

100 g of propane is completely reacted with 1000g of oxygen.The mole fraction of carbon dioxide in the resulting mixture is $x \times 10^{- 2}$ .The value of 'x' is (Rounded off to the nearest integer) is
[Atomic weight :H=1.008;C=12.00;O=16.00]

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Solution

100 g of propane = 2.27mol
1000g oxygen= 31.25 mol
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$
From the equation 1 mol of propane reacts with 5 mol of oxygen to give 3 moles of carbon dioxide and 4 moles of water
So, by stoichiometric calculations:
2.27 mol of propane will react with 11.35 mol oxygen to give 6.81 mol carbon dioxide and 9.08 mol water
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l) t = 0 2.27 31.25 00 t = \infty 0 19.9 6.81 9.08$
mole fraction of $C O_{2}$ in the final reaction mixture (heterogenous)
$X_{C O_{2}} = \frac{6.81}{19.9 + 6.81 + 9.08}$
$= 0.1902 = 19.02 \times 10^{- 2}$
Value of x is 19

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100 g of propane is completely reacted with 1000g of oxygen.The mole fraction of carbon dioxide in the resulting mixture is x×10−2.The value of 'x' is (Rounded off to the nearest integer) is [Atomic weight :H=1.008;C=12.00;O=16.00]

100 g of propane is completely reacted with 1000g of oxygen.The mole fraction of carbon dioxide in the resulting mixture is $x \times 10^{- 2}$ .The value of 'x' is (Rounded off to the nearest integer) is
[Atomic weight :H=1.008;C=12.00;O=16.00]