Calculate the standard enthalpy of formation of `CH_(3)OH(l)` from the following data:
`CH_(3)OH(l)+3/2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l), …(i), Delta_(r)H_(1)^(Θ)=-726 kJ mol^(-1)`
`C(g)+O_(2)(g) rarr CO_(2)(g), …(ii), Delta_(c )H_(2)^(Θ)=-393 kJ mol^(-1)`
`H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), ...(iii), Delta_(f)H_(3)^(Θ)=-286 kJ mol^(-1)`

Explanation

To calculate the standard enthalpy of formation of \( CH_3OH(l) \), we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Given Data:
1. Reaction (i):
\[ CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l), \quad \Delta_r H_1^\Theta = -726 \, \text{kJ/mol} \]

2. Reaction (ii):
\[ C(g) + O_2(g) \rightarrow CO_2(g), \quad \Delta_c H_2^\Theta = -393 \, \text{kJ/mol} \]

3. Reaction (iii):
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l), \quad \Delta_f H_3^\Theta = -286 \, \text{kJ/mol} \]

### Steps to Calculate the Standard Enthalpy of Formation of \( CH_3OH(l) \):

1. **Write the formation reaction for \( CH_3OH(l) \)**:
The standard enthalpy of formation is defined for the formation of one mole of a compound from its elements in their standard states. The reaction can be written as:
\[ C(s) + 2 H_2(g) + \frac{1}{2} O_2(g) \rightarrow CH_3OH(l) \]

2. **Reverse Reaction (i)**:
To find the enthalpy of formation, we need to reverse Reaction (i):
\[ CO_2(g) + 2 H_2O(l) \rightarrow CH_3OH(l) + \frac{3}{2} O_2(g) \]
The enthalpy change for this reversed reaction will be:
\[ \Delta_r H = +726 \, \text{kJ/mol} \]

3. **Use Reaction (ii)**:
The enthalpy change for Reaction (ii) is already given as:
\[ C(g) + O_2(g) \rightarrow CO_2(g), \quad \Delta_c H = -393 \, \text{kJ/mol} \]

4. **Use Reaction (iii)**:
Since we need 2 moles of \( H_2O(l) \), we will multiply Reaction (iii) by 2:
\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l), \quad \Delta_f H = 2 \times (-286) = -572 \, \text{kJ/mol} \]

5. **Combine the Reactions**:
Now we can combine the enthalpy changes:
\[ \Delta H_f^\Theta (CH_3OH) = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H_f^\Theta (CH_3OH) = (+726) + (-393) + (-572) \]
\[ \Delta H_f^\Theta (CH_3OH) = 726 - 393 - 572 \]
\[ \Delta H_f^\Theta (CH_3OH) = -239 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy of formation of \( CH_3OH(l) \) is \( -239 \, \text{kJ/mol} \).