2H\(_2\) + O\(_2\) → 2H\(_2\)O
From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm\(^3\) of hydroden and 75cm\(^3\) of oxygen are involved
- 85cm\(^3\)
-
50cm\(^3\)
- 125cm\(^3\)
- 55cm\(^3\)
The correct answer is: B
Explanation
Since 2 mol H\(_2\) reacted with 1 mol of O\(_2\)
: 50cm\(^3\) of hydrogen reacted with 25cm\(^3\) of oxygen
Remainder oxygen = Supplied amount - Reacted amount → 75cm\(^3\) - 25cm\(^3\)
= 50cm\(^3\)
There is an explanation video available .