Câu 3:
a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,25<-0,5<--------------0,25
MgO + 2HCl ---> MgCl2 + H2
=> \(\left\{{}\begin{matrix}m_{Fe}=0,25.56=14\left(g\right)\\m_{MgO}=18-14=4\left(g\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{18}.100\%=77,78\%\\\%m_{MgO}=100\%-77,78\%=22,22\%\end{matrix}\right.\)
c) \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{MgO}=0,2\left(mol\right)\)
=> \(\sum n_{HCl}=0,2+0,5=0,7\left(mol\right)\)
=> \(m_{HCl}=0,7.36,5=25,55\left(g\right)\)
d) \(C\%_{HCl}=\dfrac{25,55}{65}.100\%=39,3077\%\)
