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 step by step, we will follow the outlined procedure based on the information given in the question.

### Step 1: Identify the Components
We have a mixture of hexane (C6H14) and ethanol (C2H5OH). The total mass of the mixture is given as 10 grams.

### Step 2: Understand the Reaction
When the mixture is treated with sodium (Na), only ethanol reacts to produce hydrogen gas (H2). The balanced reaction is:
\[ \text{C}_2\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_2\text{H}_5\text{ONa} + \text{H}_2 \]

### Step 3: Calculate the Moles of Hydrogen Produced
We are given that 200 mL of hydrogen gas is produced at a temperature of 27°C (300 K) and a pressure of 760 mm Hg (1 atm). We can use the ideal gas law equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm = 1 atm
- \( V \) = volume in liters = 200 mL = 0.200 L
- \( R \) = universal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = 300 K

Rearranging for \( n \) (number of moles):
\[ n = \frac{PV}{RT} \]

### Step 4: Substitute the Values
Now we substitute the known values into the equation:
\[ n = \frac{(1 \, \text{atm}) \times (0.200 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (300 \, \text{K})} \]

### Step 5: Calculate the Number of Moles of Hydrogen
Calculating this gives:
\[ n = \frac{0.200}{24.63} \approx 0.00812 \, \text{moles of H}_2 \]

### Step 6: Relate Moles of Hydrogen to Moles of Ethanol
From the balanced reaction, we see that 1 mole of ethanol produces 1 mole of hydrogen. Therefore, the number of moles of ethanol (C2H5OH) produced is:
\[ n_{\text{C2H5OH}} = 2 \times n_{\text{H2}} = 2 \times 0.00812 = 0.01624 \, \text{moles of C2H5OH} \]

### Step 7: Calculate the Mass of Ethanol
Using the molar mass of ethanol (C2H5OH = 46 g/mol):
\[ \text{Mass of ethanol} = n \times \text{Molar mass} = 0.01624 \, \text{moles} \times 46 \, \text{g/mol} \approx 0.747 \, \text{grams} \]

### Step 8: Calculate the Percentage of Ethanol in the Mixture
To find the percentage of ethanol in the mixture:
\[ \text{Percentage of ethanol} = \left( \frac{\text{Mass of ethanol}}{\text{Total mass of mixture}} \right) \times 100 \]
\[ \text{Percentage of ethanol} = \left( \frac{0.747 \, \text{g}}{10 \, \text{g}} \right) \times 100 \approx 7.47\% \]

### Final Answer
The percentage of ethanol in the mixture is approximately **7.47%**.

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