B + dd H2SO4 loãng, dư => 1,12 lít khí H2
=> B có Fe (dư)
\(Fe+Cu\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_2+Cu\\ Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\\ Fe_{\left(dư\right)}+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe\left(dư\right)}=n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ \Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\\ \Rightarrow m_{Cu}+m_{Ag}=7,52-m_{Fe\left(dư\right)}=7,52-2,8=4,72\left(g\right)\\ Đặt:n_{Cu\left(NO_3\right)_2}=a\left(mol\right);n_{AgNO_3}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+0,5b=0,05\\64a+108b=4,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,02\end{matrix}\right.\\ \Rightarrow C_{MddCu\left(NO_3\right)_2}=\dfrac{0,04}{0,1}=0,4\left(M\right);C_{MddAgNO_3}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)
