a mixture of 10cm3 of methane and 10cm3 of enthane was sparked with an excess of oxygen. after coolin to room temp, the residual gas was passed through aq KOH. what volume of gas was absorbed by the alkali
?
Write the equations for CH4 + O2
and C2H6 + O2.
Balance them.
Look to see how many mols CO2 are formed for each mole CH4 and each mole of C2H6. Convert to cc CO2. The water vapor formed will condense and I assume the problem is neglecting any solubility of CO2 in water. Post your work if you get stuck.
i did
****2C2H6 + 7O2 -----> 4CO2 + 6H20
1 MOLE = 40
X MOLE = 10
10/40 = 0.25 MOLES
MOLE RATIO IS 2 TO 4
SO 2 * 0.25 = 0.5
0.5* 24 = 12
**** CH4 + 2O2 -----> 2H20 + CO2
1 MOLE OF CH4 = 16
X MOLE = 10
10/16 = 0.625 * 24 = 15
15 + 12 = 27
AM I RIGHT?
You seem to be equating 10 cc (a volume) with grams for a mol. Because 16 is the molar mass of CH4 but isn't 30 the molar mass C2H6 (and not 40). Anyway, you can't calculate mole this way. Try again and/or explain what you don't understand about the next step.
well, in that case, i don't understand nething abt it then. How do u find the moles then?
There is a long way and a short way.
The short way of working the problem is to recognize that volumes of gases are involved in the two equations and we can use volumes in place of moles. Usually that isn't allowed. So the short way is to see that
1 mol CH4 produces 1 mol CO2 from your balanced equation and so that 10 cc CH4 produces 10 cc CO2.
For ethane, 2 mols ethane produce 4 mols CO2 (or 1 mol ethane produces 2 mol CO2) so 10 cc ethane will produce 20 cc CO2. Then add the two volumes together for 30 cc.
The long way is to assume some numbers and use PV=nRT
I would assume 1 atm for P, volume is 0.01 L and 0.02 L respectively, R is the constant and I guess we could take room temperature as 298. Calculate mols CH4 and convert to mols CO2 produced. Calculate mols C2H4 and convert to mols CO2 produced. Add the two mols and substitute back into PV = nRT to calculate volume. You should get 0.03 L if you do it this way.
Let me know if you have any questions.
Thnx. appreciate it
i didn't get that concept but i do now.
Thnx again