Question

10ml of gaseous hydrocarbon were exploded with 70 ml(excess) of oxygen. after explosion the residual gases had a volume of 50ml which was reduced to20ml on exposure to alkali identify the hydrocarbon ( all volume measurements were made at spt)

Answer 1

Oh man.

Answer 2

what i will give the options c2h4 c4h8 c3h8 c4h10

Answer 3

the answer is c3h8 plz try to do it

Answer 4

Well, maybe we try to calculate the the molar quantities of these gasses.

Answer 5

c2h4:28
c4h8:56
c3h8:44
c4h10:58

Answer 6

ok then......

Answer 7

No, I don't know. Uh, if you can wait mayb 10 or 20 mins, my mom is coming to pick me up, and she should know (she was a chem engineer)

Answer 8

The key here is probably to look at the hydrocarbon itself I bet.

Answer 9

ok i will wait

Answer 10

Different hydrocarbons most likely have different burning reactions

Answer 11

any hydrocarbons when burned will give out c02 h20 and heat

Answer 12

Why does volume reduce on exposure to alkali?

Answer 13

i am also confused

Answer 14

plzz dont forget to ask ur mother>>

Answer 15

She's coming right now :)

Answer 16

Hopefully she didn't forget.

Answer 17

ya u r right

Answer 18

uh, 50ml is CO2

Answer 19

yes

Answer 20

plus O2 actually.
My mom's not helping -.- . Here, let me go home, and hten bother her more lol.

Answer 21

cya in 5, maybe 10 mins.

Answer 22

ok

Answer 23

Let me make a note before I leave, so I don't forget myself
CnHm+(m/4)O2=nCO2+XO2

Answer 24

I have it @shameer1 !

Answer 25

what is n , m and x here plzzz explain

Answer 26

the residual gases had a volume of 50ml which was reduced to20ml on exposure to alkali
basically, there residual gas currently has oxygen and carbon dioxide; when exposed to alkali, the carbon dioxide reacts with the alkali to make salt, and what remains is oxygen. In any case what is left is 20 ml of oxygen.

Answer 27

in other words, we can determine that 10 ml of the gaseous hydrocarbon reacted with 70-20ml of oyxgen, or 50 ml of oxygen.

Answer 28

CnHm+(n+m/4)O2=nCO2+(m/2)H2O
->n and m are unknown numbers.
CnHm+(n+m/4)O2=nCO2+(m/2)H2O
we know that we have 10 ml of hydrocarbon, and 50 ml of oxygen (this is how much reacted), and we got 30 ml of CO2

Answer 29

what is the final answer u get

Answer 30

not done :S

Answer 31

Find the vapor density of the selections first
c2h4:28 c4h8:56 c3h8:44 c4h10:58

Answer 32

@shameer1 ?

Answer 33

then what to do

Answer 34

Well, we need to find the ratio of c to h

Answer 35

@FoolForMath fool fool, helpy helpy

Answer 36

plz can u do it and show me

Answer 37

did ur mother help u

Answer 38

No, she short of ignored me, lol. I'm going to tag some more people, because I just realized that I might have to go soon, and I'm not sure if I'm even doing the problem correctly. @Diyadiya
@nbouscal
@KingGeorge

Answer 39

guys do u heve idea to solve this

Answer 40

no sorry shameer i tried but dunno :(

Answer 41

oh...no..no

Answer 42

@JFraser @shankvee

Answer 43

I'm going to repeat the problem, because it is way too far up. I will also add what I know.
10ml of gaseous hydrocarbon were exploded with 70 ml(excess) of oxygen. after explosion the residual gases had a volume of 50ml which was reduced to20ml on exposure to alkali identify the hydrocarbon ( all volume measurements were made at spt)
We know that: 20 ml is oxygen, and 30 ml is CO2 (residual gas)

Answer 44

CH being the unknown hydrocarbon.

Answer 45

CH (10 ml)+O2 (50 ml)->[CO2](30ml)+H2O

Answer 46

what does the molecular structure of a gas have to do with its density @shameer1

Answer 47

molecular mass = 2* VD

Answer 48

any idea what we can do with it and the equation?

Answer 49

VD of oxygen would be 16.
VD of CO2?

Answer 50

The answer is C3H8. Ok!

Balance the equation and do the problem the 'forward-way' (knowing 10mL of C3H8 and finding the 30mL CO2 and 20mL excess O2).

You will see the mechanism of it, and then, you can solve your problem, which is worded the 'reversed-way'.

Answer 51

we have to find the formula of hydrocarbon

Answer 52

@Vincent-Lyon.Fr
why we don have to indlude vapours of H2O to gases??

Answer 53

In standard conditions, water is liquid, so its volume is negligible.

Answer 54

you are so right AGAIN!!!!!

Answer 55

all together
CnHm + (4n+m)/4 O2 --> nCO2 + (m/2)H2O
initial 10 70 0 0
reacts/produce -10 -10*(4n+m)/4 +10n +10*(m/2)
final 0 70-10(4n+m)/4 10n 10*(m/2)

by given data you have:
Vco2=30
10n =30
n=3
and
Vo2=20 (H20 is liquid in STP)
70-10(4n+m)/4 =20 (n=3)
solve to m .....m=8
so
C3H8

Answer 56

@shameer1: have you tried doing what I suggested?

Answer 57

Try to solve that problem:

"10 mL of C3H8 are mixed with 70mL of O2. What is the residual gas volume after explosion? An alkali sucks in all the CO2 formed. What is the volume of gas left?"

Then you will see how it all works and you will be able to solve the initial problem, because you will understand the inner-mechanism behind it.

Answer 58

honestly i dint read all replies but did you manage to solve the problem?

Answer 59

can some one do this question in other way i cant understand those which are witten plzzz

Answer 60

for constant temperature & pressure:
number of moles is proportional to volume.

@shameer1
can't show any different way to do this
follow mos1635

or ask what you don't understand

Answer 61

Vco2=30 how the volume became 30

Answer 62

CO2 is produced in reaction ... but then it reacts with alkali to produce salt (since CO2 is acidic). So, Vol. CO2 = 50 - 20 = 30ml

Answer 63

what is this 70-10(4n+m)/4

Answer 64

plz explain this paxx

Answer 65

70 is initial volume of O2
minus the vol.of O2 used \(10 \times {\left( 4m+n \right)}/ 4\)

is equal to the final volume of O2 i.e. 20

Answer 66

\[\Large C_3H_8+5O_2 \rightarrow 3CO_2+4H2O\]

Answer 67

70 is initial volume of O2
minus the vol.of O2 used 10×(4m+n)/4

is equal to the excess volume of O2 i.e. 20