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

Options

A. 1 mole `H_(2)`
B. `(1)/(2) "mole" H_(2)`
C. 2 mole `H_(2)`
D. `(1)/(2) "mole H_(2) O`

Answer

To solve the problem of how much sodium reacts with methyl alcohol (methanol) and what products are formed, we can follow these steps:

### Step 1: Identify the Reaction
The reaction involves sodium (Na) reacting with methanol (CH₃OH). The balanced chemical equation for this reaction is:
\[ 2 \text{CH}_3\text{OH} + 2 \text{Na} \rightarrow 2 \text{CH}_3\text{ONa} + \text{H}_2 \]
This means that sodium methoxide (sodium methanoate) and hydrogen gas are produced.

### Step 2: Calculate the Moles of Sodium
To find out how many moles of sodium are present in 23 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of sodium (Na) is 23 g/mol. Therefore:
\[ \text{Number of moles of Na} = \frac{23 \text{ g}}{23 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the Stoichiometry of the Reaction
From the balanced equation, we see that 1 mole of sodium reacts with 1 mole of methanol to produce 1 mole of sodium methoxide and 0.5 moles of hydrogen gas.

### Step 4: Calculate the Amount of Hydrogen Produced
Since we have 1 mole of sodium, according to the stoichiometry of the reaction, it will produce:
\[ \text{Hydrogen gas produced} = 0.5 \text{ moles of H}_2 \]

### Conclusion
Thus, when 23 grams of sodium react with methyl alcohol, the products formed are:
- 1 mole of sodium methoxide (sodium methanoate)
- 0.5 moles of hydrogen gas (H₂)

### Final Answer
23 g of sodium will react with methyl alcohol to give **0.5 moles of hydrogen gas (H₂)**.

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