\(Bài.3:\\ 2< \dfrac{n_{AgNO_3}}{n_{Fe}}=\dfrac{0,025}{0,01}=2,5< 3\\PTHH:Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\left(1\right)\\ Fe\left(NO_3\right)_2+AgNO_{3\left(dư\right)}\rightarrow Fe\left(NO_3\right)_3+Ag\left(2\right)\\ n_{AgNO_3\left(2\right)}=0,025-2.0,01=0,005\left(mol\right)\\ n_{Fe\left(NO_3\right)_3}=n_{Fe\left(NO_3\right)_2\left(2\right)}=n_{AgNO_3\left(dư\right)}=0,005\left(mol\right)\\ ddY:Fe\left(NO_3\right)_2,Fe\left(NO_3\right)_3\\ n_{Fe\left(NO_3\right)_2\left(còn\right)}=0,01-0,005=0,005\left(mol\right)\\ m=m_{Fe\left(NO_3\right)_2\left(còn\right)}+m_{Fe\left(NO_3\right)_3}=0,005.180+242.0,005=2,11\left(g\right)\)
\(Bài.2:\\ n_{FeCl_2}=0,1.0,5=0,05\left(mol\right)\\ 2M+xFeCl_2\rightarrow2MCl_x+xFe\left(x:hoá.trị.M\right)\\ n_{Fe}=n_{FeCl_2}=0,05\left(mol\right)\\ n_M=\dfrac{2.0,05}{x}=\dfrac{0,1}{x}\left(mol\right)\\ m_M=0,45+0,05.56=3,25\left(g\right)\\ M_M=\dfrac{3,25}{\dfrac{0,1}{x}}=32,5x\left(\dfrac{g}{mol}\right)\\ Xét.x=1;x=2;x=3;...;thấy:x=2\left(TM\right)\Rightarrow M_M=65\left(\dfrac{g}{mol}\right)\\ Vậy:M.là.kẽm\left(Zn=65\right)\)
