23 g of sodium will react with ethyl alcohol to give

Options

A. `"1 mole of "H_(2)`
B. `"1/2 mole of "H_(2)`
C. `"1 mole of "O_(2)`
D. `"1 mole of "NaOH`

Explanation

To solve the problem of how much product will be formed when 23 g of sodium reacts with ethyl alcohol, we can follow these steps:

### Step 1: Write the Chemical Reaction
The reaction between sodium (Na) and ethyl alcohol (C2H5OH) can be represented as follows:

\[ 2 \text{C}_2\text{H}_5\text{OH} + 2 \text{Na} \rightarrow 2 \text{C}_2\text{H}_5\text{O}^- \text{Na}^+ + \text{H}_2 \]

### Step 2: Calculate the Number of Moles of Sodium
To find the number of moles of sodium, we use the formula:

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

Given:
- Mass of sodium (Na) = 23 g
- Molar mass of sodium = 23 g/mol

Calculating the number of moles:

\[ \text{Number of moles of Na} = \frac{23 \text{ g}}{23 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the Stoichiometry of the Reaction
From the balanced chemical equation, we see that 2 moles of sodium react with 2 moles of ethyl alcohol to produce 2 moles of sodium ethoxide and 1 mole of hydrogen gas. This means:

- 1 mole of sodium will react with 1 mole of ethyl alcohol.

### Step 4: Calculate the Amount of Product Formed
Since we have 1 mole of sodium, it will react with 1 mole of ethyl alcohol to produce 1 mole of sodium ethoxide (C2H5O^-Na^+) and 0.5 moles of hydrogen gas (H2).

### Conclusion
Thus, when 23 g of sodium reacts with ethyl alcohol, it will produce:
- 1 mole of sodium ethoxide
- 0.5 moles of hydrogen gas