Question

a piece of sodium metal is added to absolute ethnanal to form sodium ethoxide and hydrogen gas balance the equation and classify them ​

Answer 1

Answer:

The reaction is as follows:

Na + C2H5OH → C2H5ONa + H2

In the above reaction there are seven molecules of hydrogen on product side and six molecules on reactant side. So let us first balance the number of hydrogen.

Na + 2C2H5OH → 2C2H5ONa + H2

Now there are 12 molecules of hydrogen on both the sides. Bu there are two molecules of sodium on product side and only one molecule on reactant side. So the overall balanced reaction can be written as follows:

2Na + 2C2H5OH → 2C2H5ONa + H2

This reaction is a displacement reaction in which sodium displaces hydrogen from ethanol to form sodium ethoxide

Answer 2

Answer:

Sodium ethoxide is also known as alkoxide

Explanation:

Sodium + ethanol= sodium ethoxide+ hydrogen

2Na(s) +2C2H5OH(I) =2C2H5ONa +H2(g)