PTHH: \(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)
a) Ta có: \(n_{AgNO_3}=\dfrac{300\cdot5\%}{170}=\dfrac{3}{34}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Ag}=\dfrac{3}{34}\left(mol\right)\\n_{Cu}=\dfrac{3}{68}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Ag}=\dfrac{3}{34}\cdot108\approx9,53\left(g\right)\\m_{Cu}=\dfrac{3}{68}\cdot64\approx2,82\left(g\right)\end{matrix}\right.\)
b) Coi như p/ứ vừa đủ
Theo PTHH: \(n_{Cu\left(NO_3\right)_2}=n_{Cu}=\dfrac{3}{68}\left(mol\right)\) \(\Rightarrow m_{Cu\left(NO_3\right)_2}=\dfrac{3}{68}\cdot188\approx8,29\left(g\right)\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Cu}+m_{ddAgNO_3}-m_{Ag}=293,29\left(g\right)\)
\(\Rightarrow C\%_{Cu\left(NO_3\right)_2}=\dfrac{8,29}{293,29}\cdot100\%\approx3,46\%\)
