Question

Sodium of mass $23\,g$ will 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

We know that ethyl alcohol is an organic compound and the chemical formula of ethyl alcohol is ${C_2}{H_5}OH$ and sodium is an alkali metal with atomic number $11$ and the symbol of sodium is $Na$. Sodium is a highly reactive metal.

Complete step by step answer: First, we see the Reaction between Sodium Metal and Ethanol. Sodium reacts with ethanol gradually to give hydrogen gas and leaves a colorless solution of sodium ethoxide.
If $23\,g\left( {1\,mol} \right)$ of sodium reacts with one mole of ethyl alcohol it gives $0.5\,mol$ of hydrogen.
The balanced chemical equation is,
${C_2}{H_5}OH\left( l \right) + Na\left( s \right)\xrightarrow{{}}{C_2}{H_5}ONa\left( {aq} \right) + 1/2{H_2}\left( {aq} \right)$
Therefore, option B is correct.

Note: Let us discuss about the uses of ethanol and sodium reaction as,
To dispose of small amounts of sodium: If the sodium metal is spilled on the bench or have a small amount left over from a reaction it cannot simply disposed in the sink because sodium has the great tendency to react explosively with the water, on the other hand it reacts gently with ethanol so ethanol is hence used to dissolve small quantities of waste sodium.
To test for the -OH group in alcohols: When sodium reacts aggressively with acids and produce a salt and hydrogen, be sure that the liquid testing was neutral and be clear that there was no trace of water because sodium reacts with the -OH group in water still better than with the one in an alcohol. With those conditions, if a tiny piece of sodium is added to a neutral liquid free of water and bubbles of hydrogen produced, then the liquid is an alcohol.