Question

23g of sodium will react with methanol to give

Answer 1

When 23g of sodium will react with methanol it will give 1/2 mole of H2

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Answer 2

Answer: 23 g of sodium will react with methanol to give 1 g of hydrogen gas and 54 g of sodium methoxide .

Explanation:

$2CH_3OH+2Na\rightarrow 2CH_3O^-Na^++H_2$

Moles of sodium =$\frac{23 g}{23 g/mol}=1 mol$

According to reaction 2 moles of Na gives 2 mol of $CH_3O^-Na^+$ and 1 mol of hydrogen gas.

Then 1 mol of Na will give =1 mol of $CH_3O^-Na^+$

Then 1 mol of Na will also give = $\frac{1}{2}$ moles of $H_2$ tha is 0.5 moles of hydrogen gas.

Mass of the hydrogen gas = molecular mass of hydrogen × Moles=

=2 g/mol  × 0.5 mol= 1 g

Mass of the sodium methoxide  =

=molecular mass of $CH_3O^-Na^+$  × Moles=

=54  g/mol  × 1 mol= 54 g

23 g of sodium will react with methanol to give 1 g of hydrogen gas and 54 g of sodium methoxide .