coal gas contain 40% ethyne and 30% carbon monoxide by volume if 100 ml coal gas is mixed with 300 ml of o2 determine the final composition of mixture.
Answers
Answer:
Hydrogen=45%
methane=30%
carbon monoxide=20%
acetylene=5%
Since 100 mL of this mixture is exploded with 160 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≡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 2240022400×2×45 = 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 22400×222400×30 = 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 with2240022400×2×20 = 10 mL of O2 to give 20 mL of product CO2
22400×2 mL of CH≡CH reacts with 22400×5 mL of O2
5 mL of CH≡CH reacts with22400×522400×2×5 = 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 = 160 - 105 = 55 mL
Total volume of resulting mixture = volume of oxygen unreacted+volume of products formed.
=(55+45+90+20+15) mL
=225 mL