ta có: n_Fe2O3= 16/ 160= 0,1( mol)

PTPU

Fe₂O₃+ 6HCl\(\rightarrow\) 2FeCl₃+ 3H₂O

0,1.........0,6.................................

\(\Rightarrow\) m_HCl= 0,6. 36,5= 21.9( g)

\(\Rightarrow\) C%_HCl= \(\dfrac{21,9}{146}\). 100%= 15%

2Al + 6HCl----->2AlCl3 +3H2

x---------3x-----------x-------1,5x

Fe +2HCl----->FeCl2 +H2

y-------2y----------y------y

a)

n\(_{H2}=\)\(\frac{4,48}{22,4}=0,2mol\)

Theo bài ra ta có pt

\(\left\{{}\begin{matrix}27x+56y=5,5\\1,5x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

%m\(_{Al}=\frac{0,1.27}{5,5}.100\%=49,09\%\)

%m\(_{Fe}=100\%-49,09\%=50,91\%\)

b)Theo pthh

n\(_{HCl}=2n_{H2}=0,4\left(mol\right)\)

mddHCl =\(\frac{0,4.36,5.100}{14,6}=100\left(g\right)\)

mdd =5,5 + 100-0,4=105,1(g)

Theo pthh

n\(_{AlCl3}=n_{Al}=0,1mol\)

%m\(_{AlC_{ }l3}=\frac{0,1.98}{105,1}.100\%=9,32\%\)

Theo pthh

n\(_{FeCl2}=n_{Fe}=0,2mol\)

C%FeCl2 =\(\frac{0,2.56}{105,1}.100\%=10,66\%\)

Chúc bạn hok tốt

\(n_{Al}=x;n_{Fe}=y\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ hpt:\left\{{}\begin{matrix}27x+56y=5,5\\1,5x+y=\frac{4,48}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\\\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{0,1.27}{5,5}.100\%=49,1\left(\%\right)\\\%m_{Fe}=100-49,1=50,9\left(\%\right)\end{matrix}\right.\\ m_{ddHCl}=\frac{100.\left[36,5.\left(3x+2y\right)\right]}{14,6}=100\left(g\right)\\ C\%_M=\frac{0,1.133,5+127.0,05}{5,5+100-2.\left(1,5x+y\right)}.100\%=18,74\left(\%\right)\)

a) PTHH: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)=n_{H_2}\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)

b) Theo PTHH: \(n_{CH_3COOH}=2n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow V_{CH_3COOH}=\dfrac{0,2}{0,1}=2\left(l\right)\)

c) Theo PTHH: \(n_{\left(CH_3COO\right)_2Mg}=0,1\left(mol\right)\) \(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,1\cdot142=14,2\left(g\right)\)

*Bạn nên bổ sung thêm khối lượng riêng của dd axit 

a)nMg=2,4/24=0,1 mol

2Mg + 2CH3COOH --> 2CH3COOMg + H2

 0,1           0,1                                         0,05            mol

=> vH2 = 0,05 * 22,4 =1,12 lít

b)m CH3COOH = 0,1 * 60=6 g

c)mCH3COOMg=0,1 * 83 = 8,3 g

VCH3COOH = 0,1/0,1=1 lít

V dd sau = 2,4 + 1 - 0,05*2=3,3 l

C M = 0,1/3,3=0,03M