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Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

Mg + 2HCl --> MgCl₂ + H₂

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\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{200.18,25}{100.36,5}=1\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

          0,2<----0,4<---------------0,2

            Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,1<-----0,6

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,2.24}{0,2.24+0,1.160}.100\%=23,077\%\\\%Fe_2O_3=\dfrac{0,1.160}{0,2.24+0,1.160}.100\%=76,923\%\end{matrix}\right.\)

Lượng khí hidro thoát ra là 0,15 mol.

a) Al₂O₃ + 2NaOH \(\rightarrow\) 2NaAlO₂ + H₂O

2Al + 2NaOH + 2H₂O \(\rightarrow\) 2NaAlO₂ + 3H₂.

b) Lượng Al phản ứng bằng 2/3 số mol khí hidro thoát ra và bằng 0,1 mol.

\(\Rightarrow\) %m_nhôm=\(\dfrac{27.0,1}{12,9}\).100%\(\approx\)20,93%, %m_{nhôm oxit}\(\approx\)100%-20,93%\(\approx\)79,07%.

\(n_{H2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a) Pt : \(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2|\)

             2             2             2                 2              3

           0,2           0,2                                             0,3

            \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O|\)

                1              2                  2              1

               0,1          0,2

b) \(n_{Al}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(m_{Al2O3}=15,6-5,4=10,2\left(g\right)\)

c) Có : \(m_{Al2O3}=10,2\left(g\right)\)

\(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

\(n_{NaOH\left(tổng\right)}=0,2+0,2=0,4\left(mol\right)\)

\(V_{ddNaOH}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

 Chúc bạn học tốt

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

 x                2x                            x                     x    ( mol )

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

 y                2y                           y                      y      ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+65y=11,3\\x+y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8g\\m_{Zn}=0,1.65=6,5g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{4,8}{11,3}.100=42,47\%\\\%m_{Zn}=100\%-42,47\%=57,53\%\end{matrix}\right.\)

\(m_{CH_3COOH}=60.\left(0,2+0,1\right)=18g\)

\(C\%_{CH_3COOH}=\dfrac{18}{200}.100=9\%\)

\(\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3g\end{matrix}\right.\)

\(m_{ddspứ}=11,3+200-0,3.2=210,7g\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{28,4}{210,7}.100=13,47\%\\C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{18,3}{210,7}.100=8,68\%\end{matrix}\right.\)

Bài 2 : 

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{HCl} = 0,5.36,5 = 18,25(gam)$

b)

Bảo toàn khối lượng : 

$m_A = 22,85 + 0,25.2 - 18,25 = 5,1(gam)$

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\Rightarrow24x+65y=11,3\left(1\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(\Rightarrow x+y=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2mol\\y=0,1mol\end{matrix}\right.\)

a)\(\%m_{Mg}=\dfrac{0,2\cdot24}{11,3}\cdot100\%=42,48\%\)

\(\%m_{Zn}=100\%-42,48\%=57,52\%\)

b)\(n_{HCl}=2\left(n_{Mg}+n_{Zn}\right)=2\cdot\left(0,2+0,1\right)=0,6mol\)

\(C_{M_{HCl}}=\dfrac{0,6}{0,2}=3M\)