What volume of oxygen at S.T.P. is required to completely burn 65.0 g ethyl alcohol ?

Answer

To solve the problem of determining the volume of oxygen required to completely burn 65.0 g of ethyl alcohol (C2H5OH) at standard temperature and pressure (S.T.P.), we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethyl alcohol.

The combustion of ethyl alcohol can be represented by the following balanced equation:

\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \]

This equation shows that 1 mole of ethyl alcohol reacts with 3 moles of oxygen.

### Step 2: Calculate the molar mass of ethyl alcohol (C2H5OH).

To find the molar mass of ethyl alcohol, we add the atomic masses of all the atoms in the molecule:

- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

Adding these together:

\[ \text{Molar mass of C}_2\text{H}_5\text{OH} = 24.02 + 6.048 + 16.00 = 46.068 \text{ g/mol} \approx 46.07 \text{ g/mol} \]

### Step 3: Calculate the number of moles of ethyl alcohol in 65.0 g.

Using the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

We can calculate the number of moles of ethyl alcohol:

\[ \text{Number of moles of C}_2\text{H}_5\text{OH} = \frac{65.0 \text{ g}}{46.07 \text{ g/mol}} \approx 1.41 \text{ moles} \]

### Step 4: Determine the number of moles of oxygen required.

From the balanced equation, we know that 1 mole of ethyl alcohol requires 3 moles of oxygen. Therefore, the number of moles of oxygen required is:

\[ \text{Moles of O}_2 = 1.41 \text{ moles C}_2\text{H}_5\text{OH} \times 3 = 4.23 \text{ moles O}_2 \]

### Step 5: Calculate the volume of oxygen at S.T.P.

At standard temperature and pressure (S.T.P.), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of oxygen required can be calculated as follows:

\[ \text{Volume of O}_2 = \text{moles of O}_2 \times 22.4 \text{ L/mol} \]

\[ \text{Volume of O}_2 = 4.23 \text{ moles} \times 22.4 \text{ L/mol} \approx 94.67 \text{ L} \]

### Final Answer

The volume of oxygen required to completely burn 65.0 g of ethyl alcohol at S.T.P. is approximately **94.67 liters**.

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