Gọi \(\left\{{}\begin{matrix}n_{Cu\left(NO_3\right)_2}=a\left(mol\right)\\n_{AgNO_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{Fe\left(bđ\right)}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{Fe\left(NO_3\right)_2}=0,1.0,2=0,02\left(mol\right)\)
PTHH:
\(Fe+Cu\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_2+Cu\downarrow\)
a<------a-------------->a-------------->a
\(Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\downarrow\)
0,5b<---b---------->0,5b---------->b
\(\rightarrow m_{Fe\left(dư\right)}=\left(0,1-0,02\right).56=4,48\left(g\right)\)
=> Ta có HPT: \(\left\{{}\begin{matrix}a+0,5b=0,02\\64a+108b+4,48=7,84\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{475}\left(mol\right)\\b=\dfrac{13}{475}\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}C_{M\left(Cu\left(NO_3\right)_2\right)}=\dfrac{\dfrac{3}{475}}{0,2}=\dfrac{3}{95}M\\C_{M\left(AgNO_3\right)}=\dfrac{\dfrac{13}{475}}{0,2}=\dfrac{13}{95}M\end{matrix}\right.\)
