In the reaction
`C_(3)H_(6)(g)+nO_(2)(g)toCO_(2)(g)+H_(2)O(l)`. The ratio of the coefficients of `CO_(2)` and `H_(2)O` is

Options

A. `1:1`
B. `1:3`
C. `2:3`
D. `3:2`

Answer

To solve the question regarding the reaction \( C_3H_6(g) + nO_2(g) \rightarrow CO_2(g) + H_2O(l) \) and to find the ratio of the coefficients of \( CO_2 \) and \( H_2O \), we will follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced equation is:
\[ C_3H_6 + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Balance the carbon atoms
In the reactant \( C_3H_6 \), there are 3 carbon atoms. Therefore, we need 3 molecules of \( CO_2 \) to balance the carbon:
\[ C_3H_6 + O_2 \rightarrow 3CO_2 + H_2O \]

### Step 3: Balance the hydrogen atoms
In \( C_3H_6 \), there are 6 hydrogen atoms. To balance the hydrogen, we need 3 molecules of \( H_2O \):
\[ C_3H_6 + O_2 \rightarrow 3CO_2 + 3H_2O \]

### Step 4: Balance the oxygen atoms
Now, we need to count the oxygen atoms on the right side of the equation. We have:
- From \( 3CO_2 \): \( 3 \times 2 = 6 \) oxygen atoms
- From \( 3H_2O \): \( 3 \times 1 = 3 \) oxygen atoms
- Total oxygen on the right side = \( 6 + 3 = 9 \) oxygen atoms

To provide 9 oxygen atoms from \( O_2 \), we need:
\[ \frac{9}{2} O_2 \]
Thus, we write:
\[ C_3H_6 + \frac{9}{2} O_2 \rightarrow 3CO_2 + 3H_2O \]

### Step 5: Clear the fraction by multiplying the entire equation by 2
To eliminate the fraction, we multiply the entire equation by 2:
\[ 2C_3H_6 + 9O_2 \rightarrow 6CO_2 + 6H_2O \]

### Step 6: Identify the coefficients
From the balanced equation:
- Coefficient of \( CO_2 \) = 6
- Coefficient of \( H_2O \) = 6

### Step 7: Calculate the ratio
The ratio of the coefficients of \( CO_2 \) to \( H_2O \) is:
\[ \text{Ratio} = \frac{6}{6} = 1 \]

### Final Answer
The ratio of the coefficients of \( CO_2 \) to \( H_2O \) is \( 1:1 \).

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