\(n_{Al}=n_{Fe}=a\left(mol\right)\)\(\Rightarrow27a+56a=8,3\)
\(\Rightarrow a=0,1\left(mol\right)\)
Đặt \(n_{Cu\left(NO_3\right)_2}=x\left(mol\right);n_{AgNO_3}=y\left(mol\right)\)
Vì cho A qua HCl có khí thoát ra => X dư sau phản ứng, 2 muối hết.
Chất rắn B là Cu và Ag => \(64x+108y=28\left(1\right)\)
Ta có: \(n_{H_2}=0,05\left(mol\right)< n_{Fe}=0,1\left(mol\right)\)
Suy ra Fe dư sau phản ứng, \(n_{Fe}\left(dư\right)=0,05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\Rightarrow n_{Fe}\left(pứ\right)=0,05\left(mol\right);n_{Al}\left(pứ\right)=0,1\left(mol\right)\)
\(Al\left(0,1\right)\rightarrow Al^{+3}+3e\left(0,3\right)\)
\(Fe\left(0,05\right)\rightarrow Fe^{+2}+2e\left(0,1\right)\)
\(Cu^{+2}\left(x\right)+2e\left(2x\right)\rightarrow Cu\)
\(Ag^{+1}\left(y\right)+1e\left(y\right)\rightarrow Ag\)
Bảo toàn e => \(2x+y=0,3+0,1\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}CM_{Cu\left(NO_3\right)_2}=1\left(M\right)\\CM_{AgNO_3}=2\left(M\right)\end{matrix}\right.\)
