\(n_{AgNO_3}=\dfrac{17}{170}=0,1\left(mol\right)\\ n_{Fe}=\dfrac{4,2}{56}=0,075\left(mol\right)\\ Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\\ V\text{ì}:\dfrac{0,1}{2}< \dfrac{0,075}{1}\Rightarrow Fe.d\text{ư}\Rightarrow m_{kl}=m_{Ag}+m_{Fe\left(d\text{ư}\right)}=0,1.108+\left(0,075-\dfrac{0,1}{2}\right).56=12,2\left(g\right)\)
=> KL thanh kim loại tăng 8 gam.
