A `27 g` sample of an unknown hydrocarbon was burned in excess `O_(2)` to form `88g` of `CO_(2)` and `27g` of `H_(2)O` .What is possible molecular formula of hydrocarbon

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:-

11.2 litre of a hydrocarbon at STP produces 44.8 litre of CO_(2) at STP and 36 gm of H_(2)O during its combustion. Calculate the molecular formula of hydrocarbon.

3g of a hydrocarbons on combusion in excess of oxygen produces 8.8g of CO_(2) and 5.4g of H_(2)O . The data illustrates the law of:

1.5 g of hydrocarbon on combustion in excess of oxygen produces 4.4 g of CO_(2) and 2.7 g of H_(2)O , the data illustrates

6g of a hydrocarbon on combustion in excess of oxygen produces 17.6g of CO_(2) and 10.8g of H_(2)O . The data illustrates the law of :

How much oxygen is formed from 264 g of CO_(2) and 216 g of H_(2)O ?

How much oxygen is formed from 264 g of CO_(2) and 216 g of H_(2)O ?