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

Question

100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x × 10-2. The value of x is (Nearest integer) [Atomic weight: H =1.008 ; C =12.00 ; O =16.00 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a54151c8cf2bded0e55245e2bdef6064-.png" /> mole fraction of CO 2 in the final reaction mixture (heterogenous) X CO 2=(6.81/19.9+6.81+9.08) =0.1902=19.02 × 10-2 ⇒ 19

Q. $100 g$ of propane is completely reacted with $1000 g$ of oxygen. The mole fraction of carbon dioxide in the resulting mixture is $x \times 1 0^{- 2}$ . The value of $x$ is ______(Nearest integer)
[Atomic weight: $H = 1.008; C = 12.00; O = 16.00$ ]

mole fraction of $C O_{2}$ in the final reaction mixture (heterogenous)
$X_{C O_{2}} = \frac{6.81}{19.9 + 6.81 + 9.08}$
$= 0.1902 = 19.02 \times 1 0^{- 2}$
$\Rightarrow 19$

Q. 100g 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 ______(Nearest integer) [Atomic weight: H=1.008;C=12.00;O=16.00 ]

mole fraction of CO2​ in the final reaction mixture (heterogenous) XCO2​​=19.9+6.81+9.086.81​ =0.1902=19.02×10−2 ⇒19

Q. $100 g$ of propane is completely reacted with $1000 g$ of oxygen. The mole fraction of carbon dioxide in the resulting mixture is $x \times 1 0^{- 2}$ . The value of $x$ is ______(Nearest integer)
[Atomic weight: $H = 1.008; C = 12.00; O = 16.00$ ]

mole fraction of $C O_{2}$ in the final reaction mixture (heterogenous)
$X_{C O_{2}} = \frac{6.81}{19.9 + 6.81 + 9.08}$
$= 0.1902 = 19.02 \times 1 0^{- 2}$
$\Rightarrow 19$