Fe + 2 AgNO3 -> Fe(NO3)2 + 2Ag
nFe=\(\dfrac{11,2}{56}=0,2\left(mol\right)\)
nAgNO3=\(\dfrac{400.17\%}{170}=0,4\left(mol\right)\)
Vì 0,2.2=0,4 nên PƯ xảy ra vừa đủ
TheoPTHH ta có:
nFe=nFe(NO3)2=0,2(mol)
nAgNO3=nAg=0,4(mol)
mFe(NO3)2=0,2.180=36(g)
mdd =11,2+400-108.0,4=368(g)
C% dd Fe(NO3)2=\(\dfrac{36}{368}.100\%=9,78\%\)