Question

10 ml of gaseous hydrocarbon on combustion gives 40 ml of seo2 and 50 ml of h2o the hydrocarbon is

Answer 1

Assuming the reactants and products are in gas phase at RTP: 

CxHy+(x+y4)O2→xCO2+y2H2OCxHy+(x+y4)O2→xCO2+y2H2O

Now, gases occupying a volume at the same temperature and pressure have the same number of molecules. Hence, we can directly take out the ratio of moles that reacted:

Gas volume Ratio of the hydrocarbon, CO2CO2 and H2OH2O is 10:40:50 or simplified, 1:4:5.

Hence, the mole ratio of the hydrocarbon, CO2CO2and H2OH2O is 1:4:5.

Notice how x number of C in the hydrocarbon gives x mol of CO2CO2. That means, stoichiometrically 4 mol of CO2CO2 was given by...4 number of C! Hence,
x = 4

C4Hy+(4+y4)O2→4CO2+y2H2OC4Hy+(4+y4)O2→4CO2+y2H2O

Similarly, y number of HH gives 5 mol of H2OH2O. For each two atoms of HH, you get 1 mol of H2OH2O. That means,  5=y25=y2 or 
y = 10.

Therefore, the hydrocarbon is C4H10C4H10.

C4H10+132O2→4CO2+5H2O

Answer 2

Answer:

c4h10

Explanation:

hope its helps you