`0.2` mole sample of hydrocarbon `C_(x)H_(y)` yields after complete combustion with excess `O_(2)` gas, `0.8` mole of `CO_(2) 1.1` mole of `H_(2)O`. Hence hydrocarbon is

4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O . What is the hydrocarbon?

4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O . What is the hydrocarbon?

Four grams of hydrocarbon (C_(x)H_(y)) on complete combustion gave 12grams of CO_(2) . What is the empirical formula of the hydrocarbon ? (C = 12, H = 1)

1 mole of hydrocarbon on complete combustion give 4 mole of CO_(2) . Calculate no of carbon in hydrocarbon.

1 mole of hydrocarbon on complete combustion give 4 mole of CO_(2) . Calculate no of carbon in hydrocarbon.

When 'n' moles of CO combines with (2n+1) moles of H_2 , the hydrocarbon formed is

4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O . How many moles of oxygen is consumed per mole of hydrocarbon in the combustion?

4.4 gms of a hydrocarbon on complete combustion produced 13.2 gms of CO_(2) and 7.2 gms of H_(2)O . How many moles of oxygen is consumed per mole of hydrocarbon in the combustion?

A gaseous hydrocarbon on complete combustion gives 3.38 g of CO_(2) and 0.690 g of H_(2)O and no other product. The empirical formula of the hydrocarbon is