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 :
`ABC_3toAB+C_2`
On applying POAC for A,B&C and relating the 3 equation, we get :`(.^nABC_3)/2=(.^nAB)/2=(.^n_(c_2))/3` (`n_x` : number of moles of x)
Thus, the coefficients of `ABC_3`,AB & `C_2` in the balanced equation with be 2,2 & 3 respectively and the balanced chemical equation can be represented as :
`2ABC_3to2AB+3C_2`
Now answer the following questions :
If the weight ratio C and `O_2` presents 1:2 and both of reactants completely consume and form CO and `CO_2` and we will obtain a gasous mixture of CO and `CO_2`.What would be the weight ratio of CO and `CO_2` in mixture.

Options

A. `11:7`
B. `7:11`
C. `1:1`
D. `1:2`

Explanation

To solve the problem, we need to determine the weight ratio of carbon monoxide (CO) and carbon dioxide (CO2) formed from the complete combustion of carbon (C) and oxygen (O2) in a weight ratio of 1:2.

### Step-by-Step Solution:

1. **Identify Molar Masses:**
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Oxygen (O2) = 32 g/mol
- Molar mass of Carbon Monoxide (CO) = 28 g/mol
- Molar mass of Carbon Dioxide (CO2) = 44 g/mol

2. **Set Up the Initial Masses:**
- Given the weight ratio of C to O2 is 1:2, we can assume:
- Mass of Carbon = 100 g
- Mass of Oxygen = 200 g

3. **Calculate Moles of Reactants:**
- Moles of Carbon (C) = 100 g / 12 g/mol = 8.33 moles
- Moles of Oxygen (O2) = 200 g / 32 g/mol = 6.25 moles

4. **Reaction Consideration:**
- Let \( x \) moles of Carbon react with \( x \) moles of Oxygen to form \( x \) moles of CO2.
- The remaining Carbon will be \( 8.33 - x \) moles, which will react with \( \frac{8.33 - x}{2} \) moles of Oxygen to form \( 8.33 - x \) moles of CO.

5. **Set Up the Equation:**
- The moles of Oxygen consumed can be expressed as:
\[ \frac{8.33 - x}{2} + x = 6.25 \]
- Simplifying this gives:
\[ 8.33 - x + 2x = 12.5 \]
\[ x = 4.17 \text{ moles of CO2} \]

6. **Calculate Moles of Products:**
- Moles of CO2 formed = \( x = 4.17 \)
- Moles of CO formed = \( 8.33 - x = 8.33 - 4.17 = 4.16 \)

7. **Calculate Weight of Products:**
- Weight of CO2 = Moles of CO2 × Molar mass of CO2 = \( 4.17 \times 44 \) g = 183.48 g
- Weight of CO = Moles of CO × Molar mass of CO = \( 4.16 \times 28 \) g = 116.48 g

8. **Determine Weight Ratio:**
- Weight ratio of CO to CO2 = Weight of CO : Weight of CO2 = \( 116.48 : 183.48 \)
- Simplifying this ratio:
\[ \frac{116.48}{183.48} \approx \frac{7}{11} \]

### Final Answer:
The weight ratio of CO to CO2 in the mixture is **7:11**.