10g of mixture of hexane and ethanol is reated with Na to give 200 ml of hydrogens at `27^(@)C and 760mm` Hg pressure. What is the percentage of ethanol in the mixture.

Answer

To solve the problem, we need to determine the percentage of ethanol in a 10g mixture of hexane and ethanol that reacts with sodium to produce hydrogen gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
Ethanol reacts with sodium to produce hydrogen gas and sodium ethoxide. The reaction can be represented as:
\[ \text{C}_2\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_2\text{H}_5\text{ONa} + \text{H}_2 \]

### Step 2: Calculate Moles of Hydrogen Produced
We are given that 200 mL of hydrogen gas is produced at 27°C and 760 mm Hg. We can use the ideal gas law to find the number of moles of hydrogen produced.

1. Convert the volume of hydrogen from mL to L:
\[ V = 200 \, \text{mL} = 0.200 \, \text{L} \]

2. Convert the temperature from Celsius to Kelvin:
\[ T = 27 + 273 = 300 \, \text{K} \]

3. Convert the pressure from mm Hg to atm:
\[ P = \frac{760 \, \text{mm Hg}}{760} = 1 \, \text{atm} \]

4. Use the ideal gas equation \( PV = nRT \) to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]
Where \( R = 0.0821 \, \text{L atm/(mol K)} \).

Plugging in the values:
\[ n = \frac{(1 \, \text{atm})(0.200 \, \text{L})}{(0.0821 \, \text{L atm/(mol K)})(300 \, \text{K})} = 0.00812 \, \text{mol} \]

### Step 3: Relate Moles of Ethanol to Moles of Hydrogen
From the reaction, we see that 1 mole of ethanol produces 1 mole of hydrogen. Therefore, the moles of ethanol that reacted is also:
\[ n_{\text{C}_2\text{H}_5\text{OH}} = 0.00812 \, \text{mol} \]

### Step 4: Calculate the Mass of Ethanol
To find the mass of ethanol, we need its molar mass. The molar mass of ethanol (C2H5OH) is:
\[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (16) = 46 \, \text{g/mol} \]

Now, calculate the mass of ethanol:
\[ \text{Mass of ethanol} = n \times \text{Molar mass} = 0.00812 \, \text{mol} \times 46 \, \text{g/mol} = 0.37352 \, \text{g} \]

### Step 5: Calculate the Percentage of Ethanol in the Mixture
Now, we can find the percentage of ethanol in the original mixture:
\[ \text{Percentage of ethanol} = \left( \frac{\text{Mass of ethanol}}{\text{Total mass of mixture}} \right) \times 100 \]
\[ \text{Percentage of ethanol} = \left( \frac{0.37352 \, \text{g}}{10 \, \text{g}} \right) \times 100 = 3.7352\% \]

### Final Answer
The percentage of ethanol in the mixture is approximately **3.74%**.

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