Calculate the enthalpy of formation of ethyl alcohol from the following data :
`C_(2)H_(5)OH(l) + 3O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l), Delta_(r)H^(@) = -1368.0 kJ`
`C(s) +O_(2)(g) to CO_(2)(g), Delta_(r)H^(@) = -393.5 kJ`
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l), Delta_(r)H^(@) = -286.0 kJ`

Explanation

To calculate the enthalpy of formation of ethyl alcohol (C2H5OH), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps of the reaction.

### Given Data:
1. **Combustion of Ethyl Alcohol:**
\[ C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l), \quad \Delta_rH^\circ = -1368.0 \, \text{kJ} \]

2. **Formation of Carbon Dioxide from Carbon:**
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta_rH^\circ = -393.5 \, \text{kJ} \]

3. **Formation of Water from Hydrogen:**
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \quad \Delta_rH^\circ = -286.0 \, \text{kJ} \]

### Step-by-Step Solution:

1. **Write the formation reaction of ethyl alcohol:**
The formation of ethyl alcohol from its elements is:
\[ 2C(s) + 3H_2(g) + \frac{3}{2}O_2(g) \rightarrow C_2H_5OH(l) \]
We need to find the enthalpy change for this reaction.

2. **Manipulate the given reactions:**
We will manipulate the given reactions to derive the formation reaction of ethyl alcohol.

- **For the combustion of ethyl alcohol (reaction 1):**
We will keep it as is.

- **For the formation of carbon dioxide (reaction 2):**
We need 2 moles of carbon dioxide, so we multiply this reaction by 2:
\[ 2C(s) + 2O_2(g) \rightarrow 2CO_2(g), \quad \Delta_rH^\circ = 2 \times (-393.5) = -787.0 \, \text{kJ} \]

- **For the formation of water (reaction 3):**
We need 3 moles of water, so we multiply this reaction by 3:
\[ 3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(l), \quad \Delta_rH^\circ = 3 \times (-286.0) = -858.0 \, \text{kJ} \]

3. **Combine the modified reactions:**
Now we add the modified reactions (2 and 3) and then subtract the combustion reaction (1):
\[ (2C(s) + 2O_2(g) \rightarrow 2CO_2(g)) + (3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(l)) - (C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)) \]

This simplifies to:
\[ 2C(s) + 3H_2(g) + \frac{3}{2}O_2(g) \rightarrow C_2H_5OH(l) \]

4. **Calculate the enthalpy change:**
The total enthalpy change for the formation of ethyl alcohol is:
\[ \Delta H_f^\circ = \Delta H_{reaction 2} + \Delta H_{reaction 3} - \Delta H_{reaction 1} \]
Substituting the values:
\[ \Delta H_f^\circ = (-787.0 \, \text{kJ}) + (-858.0 \, \text{kJ}) - (-1368.0 \, \text{kJ}) \]
\[ \Delta H_f^\circ = -787.0 - 858.0 + 1368.0 \]
\[ \Delta H_f^\circ = -277.0 \, \text{kJ} \]

### Final Answer:
The enthalpy of formation of ethyl alcohol (C2H5OH) is:
\[ \Delta H_f^\circ = -277.0 \, \text{kJ} \]