10 ml of mixture of CH4, C2H4 and CO2 are exploded with excess O2. A contraction of 17ml was observed after explosion. After the treatment with KOH solution there was a 2nd contraction of 14ml. Find the % of gases in mixture
Answers
Let the volume of CH4 be a ml
Volume of C2H4 = b ml
Hence volume of CO2 = (10 - a - b) ml
After the explosion, there was contraction of 17ml which means that when the explosion takes place at that time water is in gaseous state and after the explosion water vapors converted into liquid state.Also the volume of liquid water is assumed to be zero.
Hence 2a + 2b = 17 ........(1)
Because KOH absorbs CO2 gas when the gaseous mixture is treated with KOH ,there is a further reduction in volume of about 14 ml, hence volume of CO2 in gaseous mixture is 14 ml.
10 - a - b + a + 2b = 14.. (2)
=> 10 + b = 14
=> b = (14-10)ml = 4 ml
By using this value in equation (1), we get
2a + 2b = 17
=> 2a = 17- 2 x 4 = 17-8 = 9
=>a = 9/2 = 4.5 ml
So, volume of CH4 in the mixture = 4.5 ml
Also,Volume of C2H4 in the mixture = 4 ml
That of CO2 in the mixture= (10 - 4 - 4.5) = 1.5 ml
Hope it helps!!
CH4 and C2H4 only react with O2 to produce CO2 and H2O. The CO2 present in the original mixture donot react and remains as such. Let the volume of CH4 = x ml, Volume of C2H4 = y ml, Therefore, Volume of CO2 = (10-x-y ) ml Now, contraction in volume due to explosion = 17ml and contraction of volume after treatment with KOH = 14 ml = volume of CO2 produced. The two combustion reactions are as follows: CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) (i) x ml 2x ml x ml C2H4 (g) + 3 O2 → 2CO2 (g) + 2 H2O (l) (ii) y ml 3y ml 2y ml Contraction in volume in reaction (i) x + 2x – x = 2x ml Contraction in volume in reaction (ii) y + 3y – 2y = 2y ml Total contraction in volume = 2(x+y) ml Now, 2(x+y) = 17 (iii) And Total volume of CO2 produced in reaction (i) and (ii), x +2y So, the total volume of CO2 present in the mixture = x + 2y + 10 – x- y = (y + 10) ml This volume of CO2 is absorbed in KOH Therefore, y +10 = 14 Or y = 4 ml Substituting this value in equation (iii), we get: x = 4.5 ml And 10 – x – y = 10 – 4.5 – 4 = 1.5 ml Hence, volume of CH4 in the mixture = 4.5 ml Volume of C2H4 in the mixture = 4 ml Volume of CO2 in the mixture = 1.5 ml