elementul E este in casuta 26-> Fe
mFe:mN:mO= 28 : 21 :72 aflu cu relatia niu=m/M raportul molar:
nFe;nN:nO= 28/56 : 21/14 :72/16
= 0,5:1,5:4,5 impart prin 0,5
=1 : 3 : 9--------> FeN3O9----> Fe(NO3)3------valenta III
Fe (SO3)x are masa molara M= (56gFe + 96x, gSO4)/mol
56+96x g sare,,,,,.............32x,gS
100g...................................21,05g
x=2-------> FeSO4 ... valenta II
M,azotat= niuxM= 2x(56+3x14+9x16)g=......................calculeaza
M,sulfat= niuxM= 3x(56+32+4x16)g=....................calculeaza !!!!
