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 step by step, we will follow these instructions:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of acetic acid and ethanol
1. **For Acetic Acid (CH₃COOH)**:
- Given mass = 120 g
- Molar mass of acetic acid = 60 g/mol
- Number of moles = \(\frac{\text{mass}}{\text{molar mass}} = \frac{120 \text{ g}}{60 \text{ g/mol}} = 2 \text{ moles}\)

2. **For Ethanol (C₂H₅OH)**:
- Given mass = 92 g
- Molar mass of ethanol = 46 g/mol
- Number of moles = \(\frac{\text{mass}}{\text{molar mass}} = \frac{92 \text{ g}}{46 \text{ g/mol}} = 2 \text{ moles}\)

### Step 3: Set up the equilibrium expression
The equilibrium constant \( K_c \) for the reaction is given as 4.0. The equilibrium expression is:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

### Step 4: Define changes in concentration at equilibrium
Let \( x \) be the amount of acetic acid and ethanol that reacts at equilibrium. Thus, at equilibrium:
- Concentration of acetic acid = \( 2 - x \)
- Concentration of ethanol = \( 2 - x \)
- Concentration of ethyl acetate = \( x \)
- Concentration of water = \( x \)

### Step 5: Substitute into the equilibrium expression
Substituting the concentrations into the equilibrium expression:
\[ 4 = \frac{x \cdot x}{(2 - x)(2 - x)} = \frac{x^2}{(2 - x)^2} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 4(2 - x)^2 = x^2 \]
Expanding and rearranging:
\[ 4(4 - 4x + x^2) = x^2 \]
\[ 16 - 16x + 4x^2 = x^2 \]
\[ 3x^2 - 16x + 16 = 0 \]

### Step 7: Use the quadratic formula to find \( x \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here \( a = 3, b = -16, c = 16 \)
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} \]
\[ x = \frac{16 \pm \sqrt{256 - 192}}{6} \]
\[ x = \frac{16 \pm \sqrt{64}}{6} \]
\[ x = \frac{16 \pm 8}{6} \]
This gives two possible solutions:
1. \( x = \frac{24}{6} = 4 \)
2. \( x = \frac{8}{6} = \frac{4}{3} \)

Since \( x \) cannot exceed the initial moles of reactants, we take \( x = \frac{4}{3} \).

### Step 8: Calculate the mass of ethyl acetate produced
The molar mass of ethyl acetate (CH₃COOC₂H₅) is 88 g/mol. The mass of ethyl acetate produced is:
\[ \text{mass} = \text{moles} \times \text{molar mass} = \frac{4}{3} \text{ moles} \times 88 \text{ g/mol} = \frac{352}{3} \text{ g} \approx 117.33 \text{ g} \]

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

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