Pri zagrevanju 44,5 g pirofosforne kiseline gradi se linearni polimer sastava (HPO3)nx H2O, a pri tome se oslobađa 1,71 g vode. Odrediti prosečnu molekulsku masu nagrađenog polimera.Rešenje je 276,1.
n/2H4P2O7 -------->(HPO3)n*H2O+(n-2)/2H2O
m(H4P2O7)=44.5 g
m(H2O)=1.71 g
n(H2O)=m/Mr=1.71/18=0.095 mol
n(H4P2O7)=m/Mr=44.5/178=0.25mol
0.25 : 0.095=n/2 : (n-2)/2
0.095*n/2=0.25*(n-2)/2
n*0.095*2=0.25*2*n-0.25*2*2
0.31*n=1
n=3.225 mol
M=n*Mr(HPO3)+Mr(H2O)=3.225*80+18=276
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Pirofosforna kiselina
n/2H4P2O7 -------->(HPO3)n*H2O+(n-2)/2H2O
m(H4P2O7)=44.5 g
m(H2O)=1.71 g
n(H2O)=m/Mr=1.71/18=0.095 mol
n(H4P2O7)=m/Mr=44.5/178=0.25mol
0.25 : 0.095=n/2 : (n-2)/2
0.095*n/2=0.25*(n-2)/2
n*0.095*2=0.25*2*n-0.25*2*2
0.31*n=1
n=3.225 mol
M=n*Mr(HPO3)+Mr(H2O)=3.225*80+18=276
Pirofosforna kiselina
Hvala! :)