If the weight ratio of C and O 2 present is 1 : 2 and both of reactants comple | TestHub

Passage / Comprehension

We know that balancing a chemical equation is entirely based on the law of conservation of mass. However, the concept of the Principle of Atom Conservation (POAC) can also be related to the law of conservation of mass in a chemical reaction. So POAC can also act as a technique for balancing a chemical equation. For example, for a reaction:

$A B C_{3} \to A B + C_{2}$

On applying POAC for A, B, and C and relating the three equations, we get:

$\frac{n _{A B C_{3}}}{2} = \frac{n _{A B}}{2} = \frac{n _{C_{2}}}{3}$ (n_x : number of moles of X)

Thus, the coefficients of ABC₃, AB, and C₂ in the balanced chemical equation will be 2, 2, and 3 respectively, and the balanced chemical equation can be represented as: Now answer the following questions:

$2 A B C_{3} \to 2 A B + 3 C_{2}$

Now answer the following questions:

If the weight ratio of C and $O_{2}$ present is $1 : 2$ and both of reactants completely consume and form CO and $CO_{2}$ and we will obtain a gaseous mixture of CO and $CO_{2}$ . What would be the weight ratio of CO and $CO_{2}$ in mixture.

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Answer:

Solution:

$w C + 2 w O_{2} \to xgm CO + ygm CO_{2}$

POAC for C:

$\frac{w}{12} = \frac{x}{28} + \frac{y}{44}$

$\frac{w}{3} = \frac{x}{7} + \frac{y}{11}$ .....(i)

POAC for O

$2 \times \frac{2 w}{32} = \frac{x}{28} + 2 \times \frac{y}{44}$

$\frac{w}{4} = \frac{x}{14} + \frac{y}{11}$ ....(ii)

So now,

$\frac{3 y}{11} + \frac{3 x}{7} = \frac{4 x}{14} + \frac{4 y}{11}$

$\frac{3 x}{7} - \frac{2 x}{7} = \frac{y}{11}$

$\frac{x}{7} = \frac{y}{11}$

$\frac{x}{y} = 7 : 11$ .

Chemistry - Stoichiometry Question with Solution | TestHub

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Doubts & Discussion

JEE Advanced 2028 (Class 11)