Vì Ag không td vs H2SO4 loãng nên t có pthh:
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,15 0,15 0,15 0,15 (mol)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{Ag}=14-8,4=5,6\left(g\right)\)
\(\%m_{Fe}=\dfrac{8,4.100}{14}=60\%\)
\(\%m_{Ag}=100\%-60\%=40\%\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{14}.100\%=60\%\\\%m_{Ag}=40\%\end{matrix}\right.\)
Bạn tham khảo nhé!
