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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Feb 2008 15:27:45 IST
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How many moles of MgIn4S4 can be made from 1g each of Mg,In,S.......
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15 Feb 2008 15:41:36 IST
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hey i have just derived a small method ....plz see if it helps u
when same amount (by mass) of different elements are given and a compound is to be prepared, then multiply no. of atoms of each element by their respective atomic mass, look for the maximum value and divide the given mass of the element by this number
ex: 10g of Na and S are given and Na2S is to be prepared then multiply
at. mass of Na with no. of atoms ( 2 *23) and similarly for S ( 32*1)
max value is 2*23 = 46, so no. of moles of compound formed = 10/46
guys plz...... correct me if wrong
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15 Feb 2008 16:35:59 IST
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No,dude,this is not the right answer.
As per u'r example,u get 780/46 mols of Na2S if u do the sum in the usual way......!!!!!!
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15 Feb 2008 17:00:58 IST
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IS IT 0.00215 ?
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15 Feb 2008 17:12:50 IST
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FIND LIMITING agent.which is In.
accodring there will be combinning other elements ka moles.
see the ratio is 1: 4:4 (for 1 mole)
by solving for moles of indium we get0.0086
.: no.of moles of Mg = .25 * 0.0086
= 0.00215(appx)
= no.of moles of the compund(from the ratio)
plz rate.
feel free to ask any doubt.
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15 Feb 2008 17:32:08 IST
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No,dude,the answer isn't right.....
the options given r...
a) 0.000647
b)0.3
c)0.0917
d00.0087
and the right answer is c)
I just want to know how is c) the right answer and I want a explanation.....
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15 Feb 2008 20:25:45 IST
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antonyajay21 how can u get 780/46 moles by usual method.......
i mean it is completely out of logic
for one mole of Na2S 78 grams of total ingredients are required while in my example there are just 20grams(10+10), so how can u get such large no. of moles(780/46) from such a small qty. by my method i m getting the answer app. equal to savvej's answer
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16 Feb 2008 06:15:12 IST
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Sorry I did the wrong calculations in the previous step.Well now here's my logic.......
2Na + S --------->Na2S
46g 32g 78g
Now,46g --->32g
therefore,10g----> 320/46=6.96g
therefore,Na is limiting.3 g of sulphur is in exess.
now again, 32g------>78g
therefore, 7 g------->(78*7)/32=17.06 grams of Na2S.
Now, 78g------>1mol
therefore, 17.06g------>0.218 mols of Na2S.
What doubt I have in my question is there r three elements given and wich one is limiting the other and perfrm the stoichiometric calculations....
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16 Feb 2008 06:17:33 IST
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Even u'r shortcut method was right.But I am asking this question on the board point of view......
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