1.

2M + 2xHCl \(\rightarrow\)2MCl_x + xH₂

n_H2=\(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PTHH ta có:

n_M=\(\dfrac{2}{x}\)n_H2=\(\dfrac{0,6}{x}\)

M_M=\(\dfrac{5,4}{\dfrac{0,6}{x}}=9x\)

Với x=3 thì M_M=27

Vậy M là Al

Mg + 2 H2SO4 (đ) -to-> MgSO4 + SO2 + 2 H2O

x_________2x__________________x(mol)

2 Fe + 6 H2SO4(đ) -to-> Fe2(SO4)3 + 3 SO2 + 6 H2O

y______3y_____________________1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24x+56y=18,4\\x+1,5y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)

=> mMg= 0,3.24=7,2(g)

=> %mMg= (7,2/18,4).100=39,13%

=>%mFe= 60,87%

b) nH2SO4(tổng)=2x+3y=2.0,3+3.0,2=1,2(mol)

VddH2SO4=1,2/2=0,6(l)

a) Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+24b=18,4\)  (1)

Ta có: \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Bảo toàn electron: \(3a+2b=0,6\cdot2\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2\cdot56}{18,4}\cdot100\%\approx60,87\%\\\%m_{Mg}=39,13\%\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{SO_2}+3n_{Fe_2\left(SO_4\right)_3}+n_{MgSO_4}=1,2\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{1,2}{2}=0,6\left(l\right)=600\left(ml\right)\)

2Al + 3H₂SO₄ \(\rightarrow\)Al₂(SO₄)₃ + 3H₂ (1)

Zn + H₂SO₄ \(\rightarrow\)ZnSO₄ + H₂ (2)

a;n_H2=\(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Đặt n_Al=a

n_Zn=b

Ta có:

\(\left\{{}\begin{matrix}27a+65b=11,9\\\dfrac{3}{2}a+b=0,4\end{matrix}\right.\)

=>a=0,2;b=0,1

m_Al=27.0,2=5,4(g)

%m_Al=\(\dfrac{5,4}{11,9}.100\%=45,4\%\)

%m_Zn=100-45,4=54,6%

b;Theo PTHH 1 và 2 ta có:

n_H2=n_H2SO4=0,4(mol)

V_H2SO4=\(\dfrac{0,4}{0,5}=0,8\left(lít\right)\)

\(n_{H_2}=0,4\left(mol\right)\)

\(Zn+H_2SO_4-->ZnSO_4+H_2\uparrow\)

x.........x...............................x............x

\(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\uparrow\)

y.........1,5y.......................0,5y.................1,5y

\(\left\{{}\begin{matrix}65x+27y=11,9\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%Zn=\dfrac{0,1.65}{11,9}.100\%\approx54,62\%\)

\(\%Al=100\%-54,62\%=45,38\%\)

b) nH2=nH2SO4=0,4(mol)

\(V_{ddH_2SO_4}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

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

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

0,25 ..... 0,5 ................... 0,25 (mol) 

\(m_{Mg}=0,25.24=6\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{6}{10}.100\%=60\%\\\%m_{MgO}=100\%-60\%=40\%\end{matrix}\right.\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,1 ....... 0,2 (mol) 

\(n_{HCl}=0,25+0,1=0,35\left(mol\right)\)

\(C_M\left(HCl\right)=\dfrac{0,35}{0,1}=3,5\left(M\right)\)

BT electron:

n_{e nhường} = n_{e nhận}

↔ \(\frac{14,4}{R}\cdot n=4\cdot0,1+2\cdot\frac{13,44}{22,4}\) (R là klg mol, n là hoá trị)

→ R = 9n → R là nhôm (Al)