`C_(2)H_(6)(g)+nO_(2) rarr CO_(2)(g)+H_(2)O(l)`
In this equation, the ratio of the coefficients of `CO_(2)` and `H_(2)O` is

Options

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

Explanation

To determine the ratio of the coefficients of \( CO_2 \) and \( H_2O \) in the reaction:

\[ C_2H_6(g) + nO_2 \rightarrow CO_2(g) + H_2O(l) \]

we need to balance the chemical equation. Here’s a step-by-step solution:

### Step 1: Write the unbalanced equation
The unbalanced equation is:

\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Identify the number of atoms of each element
- On the left side, we have:
- Carbon (C): 2 (from \( C_2H_6 \))
- Hydrogen (H): 6 (from \( C_2H_6 \))
- Oxygen (O): n (from \( O_2 \))

- On the right side, we have:
- Carbon (C): 1 (from \( CO_2 \))
- Hydrogen (H): 2 (from \( H_2O \))
- Oxygen (O): 1 (from \( CO_2 \)) + 1 (from \( H_2O \)) = 2

### Step 3: Balance the carbon atoms
To balance the carbon atoms, we need 2 \( CO_2 \) on the right side:

\[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \]

### Step 4: Balance the hydrogen atoms
Now, we have 6 hydrogen atoms on the left side (from \( C_2H_6 \)) and we need to balance them with water on the right side. To do this, we need 3 \( H_2O \):

\[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \]

### Step 5: Count the oxygen atoms
Now we count the oxygen atoms on the right side:
- From \( 2CO_2 \): 2 × 2 = 4 oxygen atoms
- From \( 3H_2O \): 3 × 1 = 3 oxygen atoms

Total oxygen on the right side = 4 + 3 = 7 oxygen atoms.

### Step 6: Balance the oxygen atoms
Since \( O_2 \) provides 2 oxygen atoms per molecule, we need:

\[ \frac{7}{2} O_2 = 7 \text{ oxygen atoms} \]

So, we can rewrite the equation as:

\[ C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \]

### Step 7: Clear the fraction
To eliminate the fraction, we can multiply the entire equation by 2:

\[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \]

### Step 8: Determine the ratio of coefficients
Now, we can see the coefficients:
- Coefficient of \( CO_2 \) = 4
- Coefficient of \( H_2O \) = 6

Thus, the ratio of the coefficients of \( CO_2 \) to \( H_2O \) is:

\[ \text{Ratio} = \frac{4}{6} = \frac{2}{3} \]

### Final Answer
The ratio of the coefficients of \( CO_2 \) and \( H_2O \) is \( 2:3 \).

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