Vol of CO = 50 cc and H_{2} = 50 cc [As Water gas contains CO and H_{2} in equal ratio] and O_{2} = 100 cc

CO + H_{2} + O_{2} → CO_{2} + H_{2}O

1 vol : 1 vol : 1 vol → 1 vol : 1 vol

**To calculate the amount of O _{2} used,**

CO : O_{2}

1 vol : 1 vol

50 cc : x cc

As ratio between CO and O_{2} is same so, O_{2} used is 50 cc.

Hence, remaining O_{2} = 100 – 50 = 50 cc.

To calculate the amount of CO_{2} produced,

CO : O

1 vol : 1 vol

50 cc : y cc

_{2}

Hence, CO

_{2}produced = 50 cc.

Therefore, the resultant mixture has **50 cc of O _{2} + 50 cc of CO_{2}**