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Stoichiometric Calculations

If the weight...

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 $C O$ and $C O_{2}$ in mixture?

A

11 : 7

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B

7 : 11

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C

1 : 1

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D

1 : 2

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Solution

The correct option is C 7 : 11
The molar masses of $C$ , $O_{2}$ , $C O$ and $C O_{2}$ are 12 g/mol, 32 g/mol, 28 g/mol and 44 g/mol respectively.
Let 100 g of C and 200 g of $O_{2}$ are present which corresponds to 8.33 moles of C and 6.25 moles of $O_{2}$ .
Let x moles of C react with x moles of $O_{2}$ to form x moles of $C O_{2}$ .
$8.33 - x$ moles of C will react with $\frac{8.33 - x}{2}$ moles of $O_{2}$ to form $8.33 - x$ moles of CO.
But, $\frac{8.33 - x}{2}$ moles of $O_{2}$ is also equal to $6.25 - x$ moles.
Hence, $\frac{8.33 - x}{2} = 6.25 - x$ or $x = 4.17$ .
The number of moles of $C O$ formed is $8.33 - x = 4.16$ and the number of moles of $C O_{2}$ formed is $x = 4.17$ .
Thus, equal number of moles of $C O$ and $C O_{2}$ is formed.
Hence, the ratio of the weight percent of $C O$ and $C O_{2}$ is equal to the ratio of their molar masses which is $28 : 44$ or $7 : 11$ .

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Similar questions

Q. If the weight ratio of

C

O_{2}

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

C O

C O_{2}

. We will obtain a gaseous mixture of

C O

C O_{2}

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

O_{2}

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

C O

C O_{2}

and we will obtain a gaseous mixture of

C O

C O_{2}

. What would be the weight ratio of

C O

C O_{2}

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

O_{2}

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

C O

C O_{2}

and we will obtain a gaseous mixture of

C O

C O_{2}

. What would be the weight ratio of

C O

C O_{2}

Q. By the reaction of carbon and oxygen, a mixture of

C O

C O_{2}

is obtained. The mass ratio of the mixture obtained when

20

O_{2}

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12

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Q. Equivalent weight of carbon in

C O

{C O}_{2}

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