We know that balancing of a chemical equation is entirely based on law of conservation of mass- However the concept of Principle of Atom Conservation POAC can also be related to 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 - ABC3- AB-C2 On applying POAC for A- B C and relating the 3 equations- we get- nABC32-nAB2-nC23 nx - number of moles of X Thus- the coefficients of ABC3- AB - C2 in the balanced chemical equation will be 2- 2 3 respectively and the balanced chemical equation can bbe represented as- 2ABC3- 2AB-3C2 Now answer the following question- If the weight ratio of C and O2 present is 1 - 2 and both of reactants completely consume and form CO and CO2 and we will obtain a gaseous mixture of CO and CO2- What would be the weight ratio of CO an CO2 in mixture-

Name: JEE - Grade 11 - Chemistry - States of Matter - Session 05 - W06
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is B 7 : 11C + O_2 → CO + CO_2 nbsp;w nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; 2w nbsp; nb ...

The correct option is B 7 : 11C + O_2 → CO + CO_2 nbsp;w nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; 2w nbsp; nb ...

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Standard XII

Chemistry

Mole Concept

We know that ...

Question

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to 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 & C and relating the 3 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 $A B C_{3}, A B & C_{2}$ in the balanced chemical equation will be 2, 2 & 3 respectively and the balanced chemical equation can bbe represented as:
$2 A B C_{3} \to 2 A B + 3 C_{2}$
Now answer the following question:

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

11 : 7

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7 : 11

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1 : 1

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Solution

The correct option is B 7 : 11
$C + O_{2} \to C O + C O_{2}$
w 2w x g y g

POAC for C
$\frac{w}{12} = \frac{x}{28} + \frac{y}{44}$
$\frac{w}{3} = \frac{x}{7} + \frac{y}{11}$ ...(1)

POAC for O
$2 \times \frac{2 w}{32} = \frac{x}{28} + 2 \times \frac{y}{44}$
$\frac{w}{4} = \frac{x}{14} + \frac{y}{11}$ ...(2)

So now,
$\frac{3 y}{11} + \frac{3 x}{7} = \frac{4 x}{14} + \frac{4 y}{11}$
$\frac{3 x}{7} - \frac{2 x}{7} = \frac{y}{11}$
$\frac{x}{7} = \frac{y}{11} \frac{x}{y} = 7 : 11$

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Similar questions

Q.  We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to 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 & C and relating the 3 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

A B C_{3}, A B & C_{2}

in the balanced chemical equation will be 2, 2 & 3 respectively and the balanced chemical equation can bbe represented as:

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

Now answer the following question:

If the weight ratio of

C

and

O_{2}

present is 1 : 2 and both of reactants completely consume and form

C O

and

C O_{2}

and we will obtain a gaseous mixture of

C O

and

C O_{2}

. What would be the weight ratio of

C O

C O_{2}

in mixture.

Q. Assertion :The mass and the charge are balanced on both sides of the reaction is a balanced chemical equation. Reason: The chemical reaction should be balanced so that it can follow the law of conservation of mass.

Q. If the weight ratio of

C

and

O_{2}

present is 1:2 and both of reactants completely consume and form

C O

and

C O_{2}

and we will obtain a gaseous mixture of

C O

and

C O_{2}

. What would be the weight ratio of

C O

and

C O_{2}

in mixture?

Q. If the weight ratio of

C

and

O_{2}

present is 1:2 and both of reactants completely consume and form

C O

and

C O_{2}

. We will obtain a gaseous mixture of

C O

and

C O_{2}

in the ratio:

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Mole Concept

Standard XII Chemistry

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The correct option is B 7 : 11C + O2 → CO + CO2 w 2w x g y g POAC for C w12=x28+y44 w3=x7+y11 ...(1) POAC for O 2×2w32=x28+2×y44 w4=x14+y11 ...(2) So now, 3y11+3x7=4x14+4y11 3x7−2x7=y11 x7=y11xy=7:11