PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{100}\cdot100\%=14,6\%\end{matrix}\right.\)
a) Ta có: nH2 = \(\dfrac{4,48}{22,4}\) = 0,2 (mol)
b) PTHH:
Fe + 2HCl ➝ FeCl2 + H2
c) Theo PT thì nFe = nH2 = 0,2 (mol)
=> mFe = 0,2 . 56 = 11,2 (g)
d)Theo PT thì nHCl = 2nH2 = 2 . 0,2 = 0,4 (mol)
=> Noòng độ % dd HCl là:
C% =\(\dfrac{0,4 . 36,5. 100%}{100}\) = 14,6%
