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#Post#: 792--------------------------------------------------
stoichiometry
DIR By: PIRAH SALEEM
Date: October 2, 2012, 2:08 am
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7.
of a mixture of 5 cm3 of CH4 and 5 cm3 of C2H4 ?
A.
B.
C.
D.
E.
#Post#: 797--------------------------------------------------
Re: stoichiometry
DIR By: Michel
Date: October 2, 2012, 10:30 am
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Show your attempt...then I help you from there
#Post#: 798--------------------------------------------------
Re: stoichiometry
DIR By: PIRAH SALEEM
Date: October 2, 2012, 2:15 pm
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I`ve added the volume because it`s a mixture and then applied
stoichiometry like this
44.42 dm3 requires 22.41dm3 then 10 raise to power -2 would
require 1/2 10 raise to power -2 then I`ve converted in cm3 from
dm3 N I`ve got the answer 5.....I think I was making a mistake
of calculation last time when I was solving it....thanx for
making me solve again
#Post#: 799--------------------------------------------------
Re: stoichiometry
DIR By: PIRAH SALEEM
Date: October 2, 2012, 2:21 pm
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Ahh! unfortunately I am still wrong because here in my answer
key it is`nt the correct answer :(
#Post#: 802--------------------------------------------------
Re: stoichiometry
DIR By: Michel
Date: October 2, 2012, 3:28 pm
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Okay, good to see your attempt. Lets start by you writing the
balanced equation for the reaction of both gases with Oxygen.
#Post#: 805--------------------------------------------------
Re: stoichiometry
DIR By: dhruvin patel
Date: October 4, 2012, 7:25 am
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is it 25 cm3 of oxygen
#Post#: 807--------------------------------------------------
Re: stoichiometry
DIR By: Michel
Date: October 4, 2012, 1:53 pm
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--- Quote from: dhruvin patel link ---
>
> is it 25 cm3 of oxygen
>
>
--- End Quote ---
Please read forum rules. Even if your answer is correct, don't
post it directly, but explain how you arrived at the answer.
Thanks
#Post#: 815--------------------------------------------------
Re: stoichiometry
DIR By: Chris
Date: October 5, 2012, 4:06 pm
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Balancing the two reactions
CH4 + 2O2 => CO2 + 2H2O
from gay lussac law, 1cm3 of CH4 needs 2cm3 of O2, then 5cm3 of
CH4 needs 2*5=10cm3 of O2
C2H4 + 3O2 => 2CO2 + 2H2O
1cm3 of C2H4 needs 3cm3 of O2, then 5cm3 of C2H4 needs 3*5 =
15cm3 of O2
Therefore total volume of O2 needed= 10cm3 + 15cm3
=25cm3
#Post#: 833--------------------------------------------------
Re: stoichiometry
DIR By: PIRAH SALEEM
Date: October 6, 2012, 2:56 pm
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thanks
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