When equal weights of methyl alcohol and ethyl alcohol react with excess of sodium metal the volume of ${{\text{H}}_{\text{2}}}$ liberated is more in the case of:
Answer
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Hint: We will look at both the reactions separately. Both ethanol and methanol give hydrogen when sodium metal is dropped into the solutions of both. Ethanol and methanol are just two kinds of alcohol. Ethanol, that is also known as ethyl alcohol falls with the chemical structure with two carbon atoms. While methanol, also known as methyl alcohol is composed of only one carbon atom.
Complete step by step answer:
We will first write the chemical equations involved in solving the question.
${\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + 2Na}} \to {\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa + }}{{\text{H}}_{\text{2}}}$
In the above reaction, ethanol reacts with sodium to give sodium ethoxide and hydrogen.
${\text{2C}}{{\text{H}}_{\text{3}}}{\text{OH + 2Na}} \to {\text{2C}}{{\text{H}}_{\text{3}}}{\text{ONa + }}{{\text{H}}_{\text{2}}}$
In the above shown reaction methanol reacts with sodium to give sodium methoxide and hydrogen gas respectively.
From the above two written reactions we can infer that the ${\text{2moles}}$ of ethanol gives ${\text{1mole}}$ of ${{\text{H}}_{\text{2}}}$whereas ${\text{2moles}}$ of methanol also gives ${\text{1mole}}$ of ${{\text{H}}_{\text{2}}}$ on reacting with ${\text{Na}}$.
Now, if ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ has equal weight as given in the question.
Then, number of moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH = }}\dfrac{{\text{x}}}{{{\text{32}}}}{\text{mole}}$.
Number of moles of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH = }}\dfrac{{\text{x}}}{{{\text{46}}}}{\text{mole}}$.
As, $\dfrac{{\text{x}}}{{{\text{32}}}}{\text{ > }}\dfrac{{\text{x}}}{{{\text{46}}}}$. Thus, it means that more number of moles of methanol will be there which gives more number of hydrogen gas or more volume of hydrogen gas and lesser number of moles of ethanol will therefore produce less amount of hydrogen gas than methanol.
So, option B is correct that is hydrogen will be liberated more in the case of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$.
Note: Methanol is a poisonous chemical derived through synthetic processes, whereas commercial ethanol is produced by factory fermentation of food crops. Both substances can be used as energy sources, but methanol serves mainly as a subject of research and its use as a motor fuel in the U.S. has been largely phased out.
Complete step by step answer:
We will first write the chemical equations involved in solving the question.
${\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH + 2Na}} \to {\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa + }}{{\text{H}}_{\text{2}}}$
In the above reaction, ethanol reacts with sodium to give sodium ethoxide and hydrogen.
${\text{2C}}{{\text{H}}_{\text{3}}}{\text{OH + 2Na}} \to {\text{2C}}{{\text{H}}_{\text{3}}}{\text{ONa + }}{{\text{H}}_{\text{2}}}$
In the above shown reaction methanol reacts with sodium to give sodium methoxide and hydrogen gas respectively.
From the above two written reactions we can infer that the ${\text{2moles}}$ of ethanol gives ${\text{1mole}}$ of ${{\text{H}}_{\text{2}}}$whereas ${\text{2moles}}$ of methanol also gives ${\text{1mole}}$ of ${{\text{H}}_{\text{2}}}$ on reacting with ${\text{Na}}$.
Now, if ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ has equal weight as given in the question.
Then, number of moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH = }}\dfrac{{\text{x}}}{{{\text{32}}}}{\text{mole}}$.
Number of moles of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH = }}\dfrac{{\text{x}}}{{{\text{46}}}}{\text{mole}}$.
As, $\dfrac{{\text{x}}}{{{\text{32}}}}{\text{ > }}\dfrac{{\text{x}}}{{{\text{46}}}}$. Thus, it means that more number of moles of methanol will be there which gives more number of hydrogen gas or more volume of hydrogen gas and lesser number of moles of ethanol will therefore produce less amount of hydrogen gas than methanol.
So, option B is correct that is hydrogen will be liberated more in the case of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$.
Note: Methanol is a poisonous chemical derived through synthetic processes, whereas commercial ethanol is produced by factory fermentation of food crops. Both substances can be used as energy sources, but methanol serves mainly as a subject of research and its use as a motor fuel in the U.S. has been largely phased out.
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