b). Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,2
a). nH2= \(\dfrac{4,48}{22,4}\)= 0,2(mol)
c). nFe=\(\dfrac{0,2.1}{1}\)= 0,2(mol)
mFe= n.M= 0,2 . 56 =11,2(g)
a) \(n_{HCl}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(dktc\right)\)
b) Fe + 2HCl --> FeCl2 + H2
1 2 1 1 (mol)
0,2 0,4 0,2 0,2 (mol)
c) \(m_{Fe}=n.M=0,2.56=11,2\left(g\right)\)
d) \(m_{ctHCl}=n.M=0,4.\left(1+35,5\right)=14,6\left(g\right)\)
\(C\%_{HCl}=\dfrac{m_{ctHCl}}{m_{ddHCl}}.100\%=\dfrac{14,6}{100}.100\%=14,6\%\)