500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO2 and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is

Question

500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO2 and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is (A) C4H10 (B) C5H12 (C) C2H2 (D) C5H10

Tardigrade Admin · Accepted Answer

Suppose the formula of hydrocarbon =CxHy x=(v o l u m e o f C O2/V o l u m e o f h y d r o c a r b o n)=(2500/500)=5 underset underset500 m l underset underset underset underset1 v o l .Cx Hy + (x + (y/4)) O2 = underset underset500 x m l underset underset underset undersetx v o l .x C O2 + underset underset(500 y/2) m l underset(y/2) v o l .(y/2) H2 O (500 y/2)=3000 So y = 12 Molecular formula of hydrocarbon =C5H12 .

Q. 500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of $C O_{2}$ and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is

$C_{4} H_{10}$

$C_{5} H_{12}$

$C_{2} H_{2}$

$C_{5} H_{10}$

Q. 500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO2​ and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is

C4​H10​

C5​H12​

C2​H2​

C5​H10​

Q. 500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of $C O_{2}$ and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is

$C_{4} H_{10}$

$C_{5} H_{12}$

$C_{2} H_{2}$

$C_{5} H_{10}$