23 g of sodium will react with methyl alcohol to give :

Options

A. one mole of oxygen
B. `22.4 "dm"^3` of hydrogen gas at NTP
C. 1 mole of `H_2`
D. 11.2 L of hydrogen gas at NTP

Explanation

To solve the problem of what products are formed when 23 grams of sodium react with methyl alcohol (CH3OH), let's go through the steps systematically:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are sodium (Na) and methyl alcohol (CH3OH).

2. **Write the Balanced Chemical Equation**:
- The reaction between sodium and methyl alcohol can be represented as:
\[ 2 \text{CH}_3\text{OH} + 2 \text{Na} \rightarrow 2 \text{CH}_3\text{ONa} + \text{H}_2 \]
- From the equation, we see that 2 moles of sodium react with 2 moles of methyl alcohol to produce 2 moles of sodium methylate (CH3ONa) and 1 mole of hydrogen gas (H2).

3. **Calculate Moles of Sodium**:
- The molar mass of sodium (Na) is 23 g/mol. Therefore, 23 grams of sodium corresponds to:
\[ \text{Moles of Na} = \frac{23 \text{ g}}{23 \text{ g/mol}} = 1 \text{ mole} \]

4. **Determine the Reaction Stoichiometry**:
- From the balanced equation, 1 mole of sodium reacts with 1 mole of methyl alcohol to produce 0.5 moles of hydrogen gas (H2).

5. **Calculate the Volume of Hydrogen Gas at NTP**:
- At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Since the reaction produces 0.5 moles of hydrogen gas:
\[ \text{Volume of H}_2 = 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]

6. **Conclusion**:
- Therefore, when 23 grams of sodium react with methyl alcohol, the products formed are sodium methylate (CH3ONa) and hydrogen gas (H2), with the volume of hydrogen gas produced being 11.2 liters at NTP.

### Final Answer:
The reaction of 23 grams of sodium with methyl alcohol will produce sodium methylate and 11.2 liters of hydrogen gas at NTP.