Question

23 g of sodium will react with ethyl alcohol to give:
A. One mole of hydrogen
B. One mole of oxygen
C. one mole of NaOH
D. \[\dfrac{1}{2}\] mole of hydrogen

Answer 1

Hint: The sodium metal reacts with ethyl alcohol to give a sodium ethanoate as a main product with the release of hydrogen gas. As sodium is a very reactive metal it releases heat on reacting, therefore it is an exothermic reaction. As mass of sodium is given thus mole of sodium can be calculated.

Complete step by step answer:
The sodium reacts with ethyl alcohol to give ethanol and hydrogen gas.
The reaction of sodium and ethyl alcohol is shown below.
${C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + {H_2}$
In this reaction, ethanol reacts with sodium to form sodium ethanoate and hydrogen gas.
The formula for calculating the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles.
m is the given mass
M is the molecular weight
The molecular weight of sodium is 23 g/mol
To calculate the moles of sodium, substitute the values of mass and molar mass in the above expression.
$\Rightarrow n = \dfrac{{23g}}{{23g/mol}}$
$\Rightarrow n = 1mol$
Therefore, the mole of sodium present in 23 g sodium is 1 mole.
The reaction of 1 mol sodium with ethyl alcohol is shown below.
${C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}$
In this reaction, one mole of ethanol reacts with one mole of sodium to form one mole of sodium ethanoate and half mole of hydrogen gas.
Thus, 23 g of sodium will react with ethyl alcohol to give \[\dfrac{1}{2}\] mole of hydrogen.
Therefore, the correct option is D.

Note: Make sure to balance the equation. The number of moles of component present on the left side of the reaction should be equal to the number of moles of component present on the right side of the reaction.