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Author Topic: CHEM AND BIO HELP AND TIPS HERE  (Read 356989 times)

Offline cooldude

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #120 on: November 10, 2009, 03:07:49 pm »
Quote from: @d!_†oX!© on November 10, 2009, 02:55:03 pm
ok people..here i am again...

Tungsten metal, W, is manufactured by reducing tungsten(III)oxide, WO3, whit carbon.

WO3 + 3C -----> 3CO + W

Calculate the mass of carbon needed to reduce 116g of tungsten(III)oxide.

-------------------------------------------------------------------------------------------------------------------------------
Now, I know how to do it.
We first take out the number of moles of WO3
which comes out to be 116/232 = 0.5 moles
Then, as the ratio is 1:3
Thus, the number of moles of C = 0.5*3 = 1.5 moles
Thus, the mass of carbon = 1.5*12 = 18g

Now my question,
if the ratio according to the equation is 1:3 for WO3:C
The how is it possible that for 116g of WO3 only 18g of C is required?

So, i thought maybe i have done it wrong although this is the correct method itself, but i still tried
Having the mass of WO3 as 116g and the ratio being 1:3
thus, the mass of C be 3*116 = 348g

But, the second method is wrong, isn't it?
then how are we getting 18g as the mass of Carbon required??

Please help guys!!!


dude ur first method is right, ur second method is wrong. it doesn't matter what ur ans is as long as ur working is right. and also no need to get disconcerted by ur ans, it is likely u'll get such an ans cuz wolfram has such a high Ar
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #121 on: November 10, 2009, 03:10:36 pm »
what difference does it make if it has a high Ar????
i already said it myself that the first method is the right one...but the answer has to be a valid one dude...
u cant just right 24/6=7 if your calci says so....
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #122 on: November 10, 2009, 03:40:18 pm »
GUYS WHERE ARE YOU ALL?????????????????IS THERE NOBODY TO ANSWER THIS QUUESTION OR YOU GUYS DON'T WANNA HELP?? :P :P
PLEASE GUYS HELP ME OUT!!!!!
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Offline eightAs

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #123 on: November 10, 2009, 03:52:21 pm »
Quote from: @d!_†oX!© on November 10, 2009, 03:10:36 pm
what difference does it make if it has a high Ar????
i already said it myself that the first method is the right one...but the answer has to be a valid one dude...
u cant just right 24/6=7 if your calci says so....
Dude everything depends on moles. Mole ratio is 3:1. Masses follow because of moles.
Consider this
mass required = Moles required * Ar
So it depends on both Ar and mass
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #124 on: November 10, 2009, 03:53:47 pm »
dude i understand what u are saying but tell me what will be the answer to my question...the one that i have posted above....seee...the loooong one!!!
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Offline eightAs

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #125 on: November 10, 2009, 03:55:29 pm »
Quote from: @d!_†oX!© on November 10, 2009, 02:55:03 pm
ok people..here i am again...

Tungsten metal, W, is manufactured by reducing tungsten(III)oxide, WO3, whit carbon.

WO3 + 3C -----> 3CO + W

Calculate the mass of carbon needed to reduce 116g of tungsten(III)oxide.

-------------------------------------------------------------------------------------------------------------------------------
Now, I know how to do it.
We first take out the number of moles of WO3
which comes out to be 116/232 = 0.5 moles
Then, as the ratio is 1:3
Thus, the number of moles of C = 0.5*3 = 1.5 moles
Thus, the mass of carbon = 1.5*12 = 18g

Now my question,
if the ratio according to the equation is 1:3 for WO3:C
The how is it possible that for 116g of WO3 only 18g of C is required?

So, i thought maybe i have done it wrong although this is the correct method itself, but i still tried
Having the mass of WO3 as 116g and the ratio being 1:3
thus, the mass of C be 3*116 = 348g

But, the second method is wrong, isn't it?
then how are we getting 18g as the mass of Carbon required??

Please help guys!!!

This part is wrong. Mole ratio is 3:1 not the mass ratio. Hence, the second method is wrong.
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #126 on: November 10, 2009, 03:57:14 pm »
dude i know the second part is wrong...i have already stated that....bt now tell me how can the first part be correct??
how can ther be only 18g of C to reduce 116g of WO3
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Offline ~~~~shreyapril~~~~

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #127 on: November 10, 2009, 04:02:37 pm »
now i have a meathod!!
if 1:3 rite??
1:3::116:x

so x=348 grams!!

check for the ans!!
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Offline Ghost Of Highbury

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #128 on: November 10, 2009, 04:04:48 pm »
arre shrey that answer is rong,

@adi - dude see,

232g of WO3 reacts with 36g of C rite?

so 116g with 18g of C

y do u find the answer weird, ??  check the ratio

232-36
116-18

looks quite clear to me..
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #129 on: November 10, 2009, 04:05:02 pm »
ohhhh shrey!!!u back again with ur faltu replies.....
this has already been proved wrong....this method is wrong...read the question for what exactly m asking b4 replying...

nd y do u relate everything in the world with a ratio??
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Offline ~~~~shreyapril~~~~

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #130 on: November 10, 2009, 04:06:36 pm »
Quote from: @d!_†oX!© on November 10, 2009, 04:05:02 pm
ohhhh shrey!!!u back again with ur faltu replies.....
this has already been proved wrong....this method is wrong...read the question for what exactly m asking b4 replying...

nd y do u relate everything in the world with a ratio??
i did not read your long meathod and explainations!!
By the way good ans aadi!!
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #131 on: November 10, 2009, 04:08:04 pm »
Quote from: A@di on November 10, 2009, 04:04:48 pm
arre shrey that answer is rong,

@adi - dude see,

232g of WO3 reacts with 36g of C rite?

so 116g with 18g of C

y do u find the answer weird, ??  check the ratio

232-36
116-18

looks quite clear to me..

1)why 36g of carbon???
2)m not saying the ratio is wrong, wht i am confued about is the fact that less reducing agent is required to reduce a large amount of the substance...
i mean how can only 18g of carbon be used to reduce 116gg of WO3
my question is more practical based than theoretical.
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Offline ~~~~shreyapril~~~~

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #132 on: November 10, 2009, 04:09:33 pm »
oye dude!!
it must be 36 cause 12 is atomic mass and we using 3 moles so 3*12=36!
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Offline @d!_†oX!©

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #133 on: November 10, 2009, 04:11:06 pm »
ohhohhh...okk..got the first ans...Thanks vaan...bt my second one??
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Offline Ghost Of Highbury

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Re: ~~~All Chemistry Doubts Here~~~
« Reply #134 on: November 10, 2009, 04:12:53 pm »
vaan?..where is vaan's answer??

oye dude

WO3's mol. mass - 232

3C's mol. mass - 12*3 = 36

so 232g with 36g

116 -> 18g

simple.
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