# A sample of coal gas contains 45% H2 30% ch4 20% CO and 5% C2 H2 by volume 100 cm cube of this gaseous mixture was mixed with 190 cm cube of Oxygen and exploded calculate the volume and composition of the mixture when cool to room temperature and pressure

Dear Student,

Constituents in a sample of coal gas:
hydrogen=45%
methane=30%
carbon monoxide=20%
acetylene=5%
Since 100 mL of this mixture is exploded with 190 mL of oxygen then this means that each element will react with oxygen as per the following reactions:
2H2+ O2 ------> 2H2O
CH4 + 2O2 ------> 2H2O + CO2
2CO + O2 ------> 2CO2
2CH$\equiv$CH + 5O2 ------> 4CO2 + 2H2O  (hydrocarbons always react with oxygen to give carbon dioxide and water)

1 mol of gas = 22.4 L volume at STP
Now, by the stoichiometric relations:
22400×2 mL of  H2 reacts with 22400 mL of O2
45 mL of H2 reacts with  $\frac{22400}{22400×2}×45$ = 22.5 mL of O2  to give 45 mL of H2O

22400 mL of CH4 reacts with  22400×2mL of O2
30 mL of CH4 reacts with $\frac{22400×2}{22400}×30$ = 60 mL of O2  to give 90 mL of product ( 60 mL H2O + 30 mLCO2 )

22400×2 mL of CO reacts with 22400 mL of O2
20 mL of CO reacts with$\frac{22400}{22400×2}×20$ = 10 mL of O2  to give 20 mL of product CO2

22400×2 mL of CH$\equiv$CH reacts with 22400×5 mL of O2
5 mL of CH$\equiv$CH reacts with$\frac{22400×5}{22400×2}×5$ = 12.5 mL of O2  to give 15 mL of product  ( 10 mL CO2+ 5 mL H2O )

Total volume of oxygen reacted = 22.5 + 60 + 10 + 12.5 = 105 mL
Volume of oxygen unreacted = 190 - 105 = 85 mL
Total volume of resulting mixture = volume of oxygen unreacted+volume of products formed.
=(85+45+90+20+15) mL
=255 mL