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Chemistry 15 Online
OpenStudy (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.

OpenStudy (anonymous):

@waterineyes

OpenStudy (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

OpenStudy (anonymous):

what about atoms ?

OpenStudy (anonymous):

have you copy pasted it @hanah ?

OpenStudy (hanah):

u have to find mole not atom

OpenStudy (anonymous):

i know that .. i want for atoms too !

OpenStudy (hanah):

ok

OpenStudy (anonymous):

can you explain this ?

OpenStudy (anonymous):

i had already checked the link .. only then did i come to open study

OpenStudy (anonymous):

i didn't understand that .. so can you explain ?

OpenStudy (hanah):

its just as simple just multiply 3 as given in your question with the number of mole present in the formula of ethane

OpenStudy (anonymous):

She is right.. But it seems you are not satisfied or you did not get that ?? Right??

OpenStudy (anonymous):

yeah !

OpenStudy (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 ??

OpenStudy (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 ?

OpenStudy (anonymous):

Wait go slow... I explain you everything.. Just tell me what I asked..

OpenStudy (anonymous):

2 H and 1 O

OpenStudy (anonymous):

If you look Atom wise: Then there are two atoms of H and one atom of O Right ??

OpenStudy (anonymous):

yeah !

OpenStudy (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 ??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

*is

OpenStudy (anonymous):

Okay now moving forward: This is for single \(H_2O\) What if I said : for \(2 H_2O\)

OpenStudy (anonymous):

4 H and 2 O ?

OpenStudy (anonymous):

Yes.. So 4 moles of H and 2 moles of O..

OpenStudy (anonymous):

yeah !

OpenStudy (anonymous):

Similarly in your question: you can just understand your question as : \(3(C_2H_4)\) How many C and H are there ??

OpenStudy (anonymous):

H_6...

OpenStudy (anonymous):

6 C and 12 H

OpenStudy (anonymous):

Sorry for \(C_2H_6\)..

OpenStudy (anonymous):

okay 18 H

OpenStudy (anonymous):

How ??

OpenStudy (anonymous):

Yes now you are right.

OpenStudy (anonymous):

is it just C2H6 OR 3(C2H6)

OpenStudy (anonymous):

Ethane is just : \(C_2H_6\)

OpenStudy (anonymous):

But you are given 3 moles of it : 3 Moles of \(C_2H_6\)

OpenStudy (anonymous):

So you can write it as : \(C_2H_6 + C_2H_6 + C_2H_6 = 3 C_2H_6\)

OpenStudy (anonymous):

oh okay .. understood .. so any qn like this ..i shud follow the same steps right ?

OpenStudy (anonymous):

@hanah and @waterineyes thanks :)

OpenStudy (anonymous):

Yes.. For calculating atoms in the molecule we follow the same steps..

OpenStudy (anonymous):

okay !

OpenStudy (anonymous):

Understood ?? c part you can do or not ??

OpenStudy (anonymous):

yes that i know :)

OpenStudy (anonymous):

can you see that in nana3456 you are still typing something ?

OpenStudy (anonymous):

or is it just my comp wch makes me see like that ?

OpenStudy (anonymous):

I am wondering by seeing that..

OpenStudy (anonymous):

See this site stopped working for me when I was typing that reply.. Ha ha ha.

OpenStudy (anonymous):

:P .. from the morning you are typing something and i were wondering why you wern't entering it :P

OpenStudy (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..

OpenStudy (anonymous):

its okay :)

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