anonymous
  • anonymous
In 3 moles of ethane calculate the following. (a) no of moles of C atom. (b) no of moles of H atom. (c) no. of molecules of ethane.
Chemistry
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@waterineyes
hanah
  • hanah
Answer (i) 1 mole of C2H6 contains 2 moles of carbon atoms. Number of moles of carbon atoms in 3 moles of C2H6 = 2 × 3 = 6 (ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms. Number of moles of carbon atoms in 3 moles of C2H6 = 3 × 6 = 18 (iii) 1 mole of C2H6 contains 6.023 × 10^23 molecules of ethane. Number of molecules in 3 moles of C2H6 = 3 × 6.023 × 10^23 = 18.069 × 10^23
anonymous
  • anonymous
what about atoms ?

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anonymous
  • anonymous
have you copy pasted it @hanah ?
hanah
  • hanah
u have to find mole not atom
anonymous
  • anonymous
i know that .. i want for atoms too !
hanah
  • hanah
ok
anonymous
  • anonymous
can you explain this ?
anonymous
  • anonymous
http://freeonlinehelpns.blogspot.in/2012/06/in-three-moles-of-ethane-c2h6-calculate.html
anonymous
  • anonymous
i had already checked the link .. only then did i come to open study
anonymous
  • anonymous
i didn't understand that .. so can you explain ?
hanah
  • hanah
its just as simple just multiply 3 as given in your question with the number of mole present in the formula of ethane
anonymous
  • anonymous
She is right.. But it seems you are not satisfied or you did not get that ?? Right??
anonymous
  • anonymous
yeah !
anonymous
  • anonymous
See I give you another example so that you get some understanding: See; Let suppose we have Water (already you have too) \(H_2O\) Now tell me how many H and O are there ??
anonymous
  • anonymous
does that mean whenever the qn is find the moles i should multiply the given atoms with the number of moles present ? EVERY TIME ?
anonymous
  • anonymous
Wait go slow... I explain you everything.. Just tell me what I asked..
anonymous
  • anonymous
2 H and 1 O
anonymous
  • anonymous
If you look Atom wise: Then there are two atoms of H and one atom of O Right ??
anonymous
  • anonymous
yeah !
anonymous
  • anonymous
So: For two H. there are 2 mole of H atom For one H there are 1 mole of H atom This is according to the atoms present in \(H_2O\).. Getting ??
anonymous
  • anonymous
yes
anonymous
  • anonymous
*is
anonymous
  • anonymous
Okay now moving forward: This is for single \(H_2O\) What if I said : for \(2 H_2O\)
anonymous
  • anonymous
4 H and 2 O ?
anonymous
  • anonymous
Yes.. So 4 moles of H and 2 moles of O..
anonymous
  • anonymous
yeah !
anonymous
  • anonymous
Similarly in your question: you can just understand your question as : \(3(C_2H_4)\) How many C and H are there ??
anonymous
  • anonymous
H_6...
anonymous
  • anonymous
6 C and 12 H
anonymous
  • anonymous
Sorry for \(C_2H_6\)..
anonymous
  • anonymous
okay 18 H
anonymous
  • anonymous
How ??
anonymous
  • anonymous
Yes now you are right.
anonymous
  • anonymous
is it just C2H6 OR 3(C2H6)
anonymous
  • anonymous
Ethane is just : \(C_2H_6\)
anonymous
  • anonymous
But you are given 3 moles of it : 3 Moles of \(C_2H_6\)
anonymous
  • anonymous
So you can write it as : \(C_2H_6 + C_2H_6 + C_2H_6 = 3 C_2H_6\)
anonymous
  • anonymous
oh okay .. understood .. so any qn like this ..i shud follow the same steps right ?
anonymous
  • anonymous
@hanah and @waterineyes thanks :)
anonymous
  • anonymous
Yes.. For calculating atoms in the molecule we follow the same steps..
anonymous
  • anonymous
okay !
anonymous
  • anonymous
Understood ?? c part you can do or not ??
anonymous
  • anonymous
yes that i know :)
anonymous
  • anonymous
can you see that in nana3456 you are still typing something ?
anonymous
  • anonymous
or is it just my comp wch makes me see like that ?
anonymous
  • anonymous
I am wondering by seeing that..
anonymous
  • anonymous
See this site stopped working for me when I was typing that reply.. Ha ha ha.
anonymous
  • anonymous
:P .. from the morning you are typing something and i were wondering why you wern't entering it :P
anonymous
  • anonymous
At that time I was replying to Nana and this site did not work properly for me then.. Sorry I could not reply to you..
anonymous
  • anonymous
its okay :)

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