2CO + O\(_2\) → 2CO\(_2\)

From the equation, we see that 2 volumes of CO react with 1 volume of O\(_2\) to produce 2 volumes of CO\(_2\).

Given:

Volume of CO = 50 cm³ 

Volume of O\(_2\) = 100 cm³ 

First, let's find the limiting reactant:

CO: 50 cm³ / 2 = 25 cm³ of O\(_2\) required (based on the stoichiometry)

However, we were told that 50cm\(^3\) of carbon(II)oxide was exploded in 100cm\(^3\) of oxygen, which means Oxygen is in excess (100 cm³ ), then CO is the limiting reactant, it will be completely consumed.

Now, let's calculate the volume of CO\(_2\)` produced:

50 cm³ of CO produces 50 cm³ of CO\(_2\) i.e  (1:1 ratio)

The resulting mixture will contain 50 cm³ of CO\(_2\) and the remaining O\(_2\) (100 cm³ - 25 cm³ ) = 75 cm³

Total volume of the resulting mixture = Volume of product (CO\(_2\)) + Volume of the remaining O\(_2\)

 = 50 cm\(^3\) +  75 cm\(^3\)

 = 125 cm\(^3\)

So, the volume of the resulting mixture at the end of the reaction is 125 cm³.