The combustion of carbon(II)oxide in oxygen can be represented by equation:
2CO + O\(_2\) → 2CO\(_2\)
Calculate the volume of the resulting mixture at the end of the reaction, if 50cm\(^3\) of carbon(ii)oxide was exploded in 100cm\(^3\) of oxygen
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.
There is an explanation video available below.
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