Question

23g of sodium react with ethyl alcohol to give:
(a) 1 mol of ${{H}_{2}}$
(b) $\dfrac{1}{2}$ mol of ${{H}_{2}}$
(c) 1 mol of ${{o}_{2}}$
(d) 1 mol of NaOH

Answer 1

To solve this question we have to calculate the number of moles of sodium and then use the stoichiometric calculation to find the mol of product.

Complete step by step answer:
The reaction takes place when sodium react with ethyl alcohol is mentioned below:
$Na+{{C}_{2}}{{H}_{5}}OH\to {{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}$
From the above reaction it is clear that sodium reacts with ethyl alcohol to form sodium ethoxide and hydrogen.
The balanced chemical reaction is:
$2Na+2{{C}_{2}}{{H}_{5}}OH\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}$
2 moles of sodium reacts with 2 moles of ethanol or ethyl alcohol to give 2 moles of sodium ethoxide and 1 mol of hydrogen.
Given mass of sodium = 23g
Molar mass of sodium= 23 g
Number of mol is calculated by ratio of given mass to the molar mass.
Number of mol of sodium = $\dfrac{\text{given mass}}{\text{molar mass}}$
After putting the value we get,
Number of mol of sodium = 1 mol
According to the balanced reaction- 2 moles of sodium reacts with 2 moles of ethanol or ethyl alcohol to give 2 moles of sodium ethoxide and 1 mol of hydrogen.
So, 1 mol of sodium will react with 1 mol of ethyl alcohol and produce 1 mol of sodium ethoxide.
We know that,
2 moles of sodium produce 1 mol of hydrogen, so 1 mol of sodium produces 0.5 mol of hydrogen.
Hence the correct answer is 0.5 mol of ${{H}_{2}}$ or $\dfrac{1}{2}$ mol of ${{H}_{2}}$ i.e. option (B).

Note: It is very important to balance the chemical reaction before solving such types of questions. If the chemical reaction is not balanced properly then there will always be error in the solution.