Procenat elemenata

Submitted by Biologist on Wed, 02/12/2014 - 16:15
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15,5M HnO3 ima gustocu 1,409 g/ml.
Izracunati procentni sastav ( Ar: H=1,008; N=14,00;O=15,99949)... Bila bih vamo jako zahvalna na pomoci... :*

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milicaradenkovic - Wed, 02/12/2014 - 16:59

c(HNO3)=15.5 M=15.5 mol/l

V(HNO3)=1 l

mr=V*ro=1.409*1000=1409 g

n(HNO3)=c(HNO3)*V=15.5*1=15.5 mol

ms=n(HNO3)*Mr=63.00647*15.5=976.6 g

w(HNO3)=ms/mr=976.6/1409=0.693= 69.3 %

 

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Biologist - Wed, 02/12/2014 - 17:09

Ja se izvinjavam nisam potpuno napisala ali trazi se procentni sastav svakog elementa

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milicaradenkovic - Wed, 02/12/2014 - 17:54

976.6 g---------------------63.00647 g/mol

x g-----------------------------------3*15.99949 g/mol

x=976.6*3*15.99949/63.00647=743.98 g O

w(O)=m(O)/m(HNO3)=743.98/976.6=0.762=76.2 %

976.6 g-------------------------63.00647 g/mol

x g----------------------------------14 g/mol

x=14*976.6/63.00647=217 g N

w(N)=m(N)/m(HNO3)=217/976.6=0.222=22.2 %

976.6 g----------------------------63.00647 g/mol

x g--------------------------------------1.008 g/mol

x=976.6*1.008/63.00647=15.62 g H

w(H)=m(H)/m(HNO3)=15.62/976.6=0.016=1.6 %

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milicaradenkovic - Wed, 02/12/2014 - 18:59

Ja mislim da je greska u resenju jer kad podelis molarne mase elemenata i kiseline dobices isto resenje kao i ja sto sam dobila.