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Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

PTHH: Fe₂O₃ + 6HNO₃ ---> 2Fe(NO₃)₃ + 3H₂O

Theo PT: \(n_{HNO_3}=6.n_{Fe_2O_3}=6.0,3=1,8\left(mol\right)\)

=> \(m_{HNO_3}=1,8.63=133,4\left(g\right)\)

Ta có: \(\dfrac{133,4}{m_{dd_{HNO_3}}}.100\%=12,6\%\)

=> \(m_{dd_{HNO_3}}\approx1059\left(g\right)\)

=> \(m_{dd_{Fe\left(NO_3\right)_3}}=48+1059=1107\left(g\right)\)

Theo PT: \(n_{Fe\left(NO_3\right)_3}=2.n_{Fe_2O_3}=2.0,3=0,6\left(mol\right)\)

=> \(m_{Fe\left(NO_3\right)_3}=0,6.242=145,2\left(g\right)\)

=> \(C\%_{Fe\left(NO_3\right)_3}=\dfrac{145,2}{1107}.100\%\approx13,12\%\)

\(m_{ct}=\dfrac{4.100}{100}=4\left(g\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

           1            1            1            1

         0,1          0,1

\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

20ml = 0,02l

\(C_{M_{ddHCl}}=\dfrac{0,1}{0,02}=5\left(M\right)\)

 Chúc bạn học tốt

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂.

Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,25=\dfrac{1}{6}\left(mol\right)\)

=> \(a=m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)

Theo PT: n_HCl = \(3.n_{Al}=3.\dfrac{1}{6}=0,5\)(mol)

=> m_HCl = 0,5 . 36,5 = 18,25(g)

=> \(C\%_{HCl}=\dfrac{m_{ct_{HCl}}}{m_{dd_{HCl}}}.100\%=\dfrac{18,25}{400}.100\%=4,5625\%\) 

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:    0,5/3     0,5                       0,25

\(m_{Al}=\dfrac{0,5}{3}.27=4,5\left(g\right)\)

\(C\%_{ddHCl}=\dfrac{0,5.36,5.100\%}{400}=4,5625\%\)

a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:       x                                  1,5x

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:       y                                 y

Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)

c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)

d) 

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,2     0,6          0,2       0,3

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      0,3     0,6        0,3       0,3

\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)

e) m_{dd sau pứ} = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)

\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)