Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
Gọi \(n_{Cu}pư=x\left(mol\right)\)
\(\Rightarrow m_{Cu}pư=64x\left(g\right)\)
Theo PT: \(n_{Ag}=2n_{Cu}pư=2x\left(mol\right)\)
\(\Rightarrow m_{Ag}=108\times2x=216x\left(g\right)\)
Ta có: \(m_{Ag}-m_{Cu}pư=2,28\)
\(\Leftrightarrow216x-64x=2,28\)
\(\Leftrightarrow152x=2,28\)
\(\Leftrightarrow x=0,015\left(mol\right)\)
Vậy \(n_{Cu}pư=0,015\left(mol\right)\)
\(\Rightarrow n_{Ag}=2\times0,015=0,03\left(mol\right)\)
\(\Rightarrow m_{Ag}=0,03\times108=3,24\left(g\right)\)
b) Theo pT: \(n_{Cu\left(NO_3\right)_2}=n_{Cu}pư=0,015\left(mol\right)\)
\(\Rightarrow C_{M_{Cu\left(NO_3\right)_2}}=\dfrac{0,015}{0,3}=0,05\left(M\right)\)
