500 ml of a gaseous hydrocarbon when burnt in excess of `O_(2)` gave 2.0litres of `CO_(2)` and 2.5 litres of water vapours under same conditions. molecular formula of the hydrocarbon is:-

10ml of a gaseous hydrocarbon on combustion gives 40ml of CO_(2) and 50ml of H_(2)O vapour under the same conditions. The hydrocarbon is

500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO_2 and 3 litres of water vapours. All volume being measured at the same temperature and pressure. The formula of the hydrocarbon is :

500mL of a hydrocarbon gas burnt in excess of oxygen yielded 2500 mL of Co_2 and 3.0 liter of water vapour (all volumes measured at the same temperature and pressure). The formula of the hydrocarbon is :

The volume of CO_(2) formed when 1 litre of O_(2) reacted with 2 lit of CO under the same condition is

10 ml of gaseous hydrocarbon on combution gives 20ml of CO_(2) and 30ml of H_(2)O(g) . The hydrocarbon is :-