What is the volume of oxygen required to burn complete 45 cm\(^3\) of methane at…

Explanation

CH\(_4\)(g) + 2O\(_2\)(g) → 2H\(_2\)O(g) + CO\(_2\)(g)

First, ensure the chemical equation is balanced, stoichiometrically.

i.e CH\(_4\) reacts with  2O\(_2\) in ratio 1 : 2
 

If          1 mole of CH\(_4\) reacts with 2 moles of O\(_2\)

Then   45 cm\(^3\) CH\(_4\) react with X cm\(^3\) O\(_2\)

i.e     1cm\(^3\) = 2cm\(^3\)

        45cm\(^3\)= X cm\(^3\)

X = 45 x 2 = 90 cm\(^3\)