The equilibrium constant for the reaction
`CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O`
is `4.0` at `25^(@)C`. Calculate the weight of ethyl acetate that will be obtained when `120g` of acetic acid are reacted with `92 g` of alcohol.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of acetic acid (CH₃COOH)
1. Given mass of acetic acid = 120 g
2. Molecular weight of acetic acid (CH₃COOH) = 60 g/mol
3. Number of moles of acetic acid = mass / molecular weight = 120 g / 60 g/mol = 2 moles

### Step 2: Calculate the number of moles of ethanol (C₂H₅OH)
1. Given mass of ethanol = 92 g
2. Molecular weight of ethanol (C₂H₅OH) = 46 g/mol
3. Number of moles of ethanol = mass / molecular weight = 92 g / 46 g/mol = 2 moles

### Step 3: Set up the equilibrium expression
The reaction is:
\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

Let \( x \) be the number of moles of ethyl acetate (CH₃COOC₂H₅) formed at equilibrium.

- Initial moles:
- CH₃COOH = 2
- C₂H₅OH = 2
- CH₃COOC₂H₅ = 0
- H₂O = 0

- Change in moles:
- CH₃COOH = 2 - x
- C₂H₅OH = 2 - x
- CH₃COOC₂H₅ = x
- H₂O = x

### Step 4: Write the equilibrium constant expression
The equilibrium constant \( K_c \) is given as 4.0. The expression for \( K_c \) is:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

Substituting the concentrations:
\[ K_c = \frac{\left(\frac{x}{V}\right)\left(\frac{x}{V}\right)}{\left(\frac{2-x}{V}\right)\left(\frac{2-x}{V}\right)} \]

This simplifies to:
\[ K_c = \frac{x^2}{(2-x)^2} \]

### Step 5: Solve for \( x \)
Setting \( K_c = 4 \):
\[ 4 = \frac{x^2}{(2-x)^2} \]

Taking the square root of both sides:
\[ 2 = \frac{x}{2-x} \]

Cross-multiplying gives:
\[ 2(2-x) = x \]
\[ 4 - 2x = x \]
\[ 4 = 3x \]
\[ x = \frac{4}{3} \text{ moles} \]

### Step 6: Calculate the weight of ethyl acetate
1. Molecular weight of ethyl acetate (CH₃COOC₂H₅) = 88 g/mol
2. Weight of ethyl acetate = number of moles × molecular weight
\[ \text{Weight} = \frac{4}{3} \text{ moles} \times 88 \text{ g/mol} \]
\[ \text{Weight} = \frac{352}{3} \text{ g} \approx 117.33 \text{ g} \]

### Final Answer
The weight of ethyl acetate obtained is approximately 117.33 g.

---