Okay let me help,Hello everyone!
could anyone please explain to me how the answers are begin obtained in the following qns.
qn6.-A
qn9.-B (is there a shorter method except trial and error)
qn28.-C
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Question 1 :
When you're completely combusting an alkene you would get CO2 and H20.
If you take this example: C2H4+3O2=2CO2+2H20
As you can see, we had ethene(2 carbons) so we had to write (2)CO2 to balance the eqn, if you get the pattern i'm saying now you can solve it easily,
we have 400 dm3 of CO2 formed, therefore, since v=24n, then n=16.66... and since they said 2 moles of the alkenes this means (2CnH2n) so 2n=16 since we balance the eqn, therefore using math, 2n=16, n=8. so you know now its 2(C8H16)
Question 2-
There's no need to guess or try, using oxidation balancing method
MnO4-2 > MnO4-, the oxidation no difference is 1e since in MnO4-2 it's +6, and in MnO4- it's +7
MnO4-2 > MnO2 , the oxidtion number difference is 2e since in MnO2 it's +4 so cross multiplying to make the umber of electrons equal,
2MnO4-2> 2MnO4- and Mn04-2> MnO2
Adding these 2 half-equations together you'll get 2MnO4-2 +MnO4-2 +H2O> 2MnO4- +MnO2+ OH-
so totally you have 3MnO4-2 + H20 > 2MnO4- +MnO2 +OH- (did not balance the Hydrogen yet )
so its B .. revise chp 6 to make sure I'm right.
Question 3-
KMnO4- is an oxidising agent, and ethanol is a primary alcohol.
Heating a primary alcohol under reflux *for an hour* will mean I'm producing a carboxylic acid since I'm not distilling the aldehyde off and allowing it to further oxidise,
so this means it'll produce C2H4O2 OR CH3COOH
now let's find the number of moles of ethanol used, since n=M/Mr then n =2.3/46= 0.05,
since the yield is 60% this means that only 60% of 0.05 moles fo carboxylic acid is produced,
therefore, 0.6*0.05= 0.03.
To find the mass of ethanoic acid produced, we say Mr*n=Mass so 60*0.03 =1.8 g