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: \(m_{dd\left(sau.p/ứ\right)}=m_{Cu}+m_{ddAgNO_3}-m_{Ag}=293,29\left(g\right)\)

\(\Rightarrow C\%_{Cu\left(NO_3\right)_2}=\dfrac{8,29}{293,29}\cdot100\%\approx3,46\%\)

\(\Delta_m=2,28\left(g\right)=m_{Ag}-m_{Cu\text{ p/ứ}}\left(1\right)\\ PTHH:Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\\ \Rightarrow n_{Cu}=\dfrac{1}{2}n_{AgNO_3}\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow108n_{AgNO_3}-\dfrac{1}{2}n_{AgNO_3}\cdot64=2,28\left(g\right)\\ \Rightarrow n_{AgNO_3}=0,03\left(mol\right)\\ m_{dd_{AgNO_3}}=1,14\cdot60=68,4\left(g\right)\\ n_{Cu}=n_{CuNO_3}=0,015\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{CuNO_3}}=0,015\cdot188=2,82\left(g\right)\\m_{Cu}=0015\cdot64=0,96\left(g\right)\end{matrix}\right.\\ n_{Ag}=0,03\left(mol\right)\\ \Rightarrow m_{Ag}=0,03\cdot108=3,24\left(g\right)\)

\(\Rightarrow m_{dd_{CuNO_3}}=0,96+68,4-3,24=66,12\left(g\right)\\ \Rightarrow C\%_{CuNO_3}=\dfrac{2,82}{66,12}\cdot100\%\approx4,26\%\)

Bạn xem lại nồng độ % của dd AgNO₃

PT: \(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)

a, m _{AgNO3 (pư)} = 250.17%.6% = 2,55 (g)

\(\Rightarrow n_{AgNO_3\left(pư\right)}=\dfrac{2,55}{170}=0,015\left(mol\right)\)

Theo PT: n_{Cu (pư)} = 1/2n_AgNO3 = 0,0075 (mol)

n_Ag = n_AgNO3 = 0,015 (mol)

⇒ m _{vật lấy ra} = 50 - m_{Cu (pư)} - m_Ag = 51,14 (g)

b, Ta có: m _{dd sau pư} = 0,0075.64 + 250 - 0,015.108 = 248,86 (g)

Theo PT: n_Cu(NO3)2 = 1/2n_AgNO3 = 0,0075 (mol)

\(\Rightarrow C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,0075.188}{248,86}.100\%\approx0,57\%\)

\(C\%_{AgNO_3}=\dfrac{250.6\%-2,55}{248,86}.100\%\approx5\%\)

Gọi \(n_{Al,pư}=a\left(a>0\right)\)

\(Al+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3Ag\\ \Rightarrow m_{\uparrow}=108.3a-27a=28,5-27\\ \Rightarrow a=\dfrac{1}{198}mol\\ n_{Ag}=\dfrac{1}{198}\cdot3=\dfrac{1}{66}mol\\ m_{Ag}=\dfrac{1}{66}\cdot108=\dfrac{18}{11}=1,64g\)