Question

Câu 3

Answer 1

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) \(C\%_{HCl}=\dfrac{25,55}{65}.100\%=39,3077\%\)

Answer 2

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