20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide

[By Lussac’s law]2CO + O_{2} → 3CO_{2}

2 vol : 1 vol → 2 vol

**To calculate the amount of CO _{2} produced,**

CO : CO_{2}

2 : 2

10 : x

⇒ x = 2 × 10\over 2 = 10 ml

Therefore, CO_{2} produced is 10 ml

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

CO : O_{2}

2 vol : 1 vol

10 ml : x ml

⇒ x = 10 × 1\over 2 = 5 ml

From relation,

2H_{2} + O_{2} → 2 H_{2}O

2 vol : 1 vol → 2 vol

To calculate the amount of O_{2} used,

H_{2 } : O_{2}

2 vol : 1 vol

20 ml : x ml

⇒ x = 20 × 1\over 2 = 10 ml

Therefore, total vol. of oxygen used = 10 + 5 = 15 ml

Hence, oxygen left = 20 – 15 = 5 ml

Therefore, **oxygen left is 5 ml and CO _{2} produced is 10 ml.**