Ta có: \(n_{H_2}=\dfrac{4,2}{22,4}=0,1875\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

__0,125__0,375___________0,1875 (mol)

\(\Rightarrow m_{Al}=0,125.27=3,375\left(g\right)\)

\(V_{HCl}=\dfrac{0,375}{3}=0,125\left(l\right)=125\left(ml\right)\)

Bạn tham khảo nhé!

n H2 = 4,2/22,4 = 3/16 mol

2Al + 6HCl $\to$ 2AlCl3 + 3H2

Theo PTHH :

n HCl = 2n H2 = 3/8 mol => V dd HCl = (3/8) / 3 = 0,125M

n Al = 2/3 n H2 = 0,125(mol) => m = 0,125.27  = 3,375(gam)

Đề thiếu dữ kiện

V₁ = 200 ml = 0,2 l

ta có :
\(\dfrac{V_1}{V_2}\)= \(\dfrac{C_2-C}{C-C_1}\)<=>\(\dfrac{0,2}{V_2}\) = \(\dfrac{2-1,5}{1,5-0,5}\)

=>V₂ = 0,4 (l) = 400 ml

\(2Al+6HCl\rightarrow2AlCl3+3H_2\)

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

\(n_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

\(n_{HCl}=2n_{H2}=0,4.2=0,8\left(mol\right)\)

BTKL: m kim loại+ mHCl= m muối+mH2

\(\rightarrow7,8+0,8.36,5=m_{muoi}+0,4.2\)

\(\rightarrow m_{muoi}=36,2\left(g\right)\)

Câu b) là tính nồng độ trong X chứ :v

a) 

PTHH: Mg + 2HCl --> MgCl₂ + H₂

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

Theo PTHH: \(n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.\left(0,4.2\right)=0,4\left(mol\right)\)

=> V = 0,4.22,4 = 8,96 (l)

b) 

Gọi số mol Mg, Al là a, b (mol)

=> 24a + 27b = 7,8 (1)

Theo PTHH: n_H2 = a + 1,5b = 0,4 (2)

(1)(2) => a = 0,1 (mol); b = 0,2(mol)

=> \(\left\{{}\begin{matrix}n_{MgCl_2}=0,1\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,4}=0,5M\\C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\end{matrix}\right.\)

Câu b) là tính nồng độ trong X chứ :v

a) 

PTHH: Mg + 2HCl --> MgCl₂ + H₂

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

Theo PTHH: ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩CM(AlCl3)=0,20,4=0,5MCM(MgCl2)=0,10,4=0,25M

Theo đề, ta có: \(n_{H_2}=\dfrac{4,368}{22,4}=0,195\left(mol\right)\)

Gọi \(x,y\) lần lượt là số mol của \(Mg,Al\) phản ứng trong hỗn hợp ban đầu.

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2 \left(1\right)\)

Số mol: \(x---------->x\)

\(2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(2\right)\)

Số mol: \(y------------>\dfrac{3}{2}y\)

Theo đề, ta có hệ: \(\left\{{}\begin{matrix}24x+27y=3,87\\x+\dfrac{3}{2}y=0,195\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0,06\left(mol\right)\\0,09\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{Mg}=0,06.24=1,44\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{1,44}{3,87}=37,2\%\)

\(\Rightarrow\%m_{Al}=100\%-37,2\%=62,8\%\)

a) Gọi số mol Mg, Al là a, b (mol)

=> 24a + 27b = 1,98 (1)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a--->2a-------->a---->a

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

              b---->3b------->b----->1,5b

=> \(n_{H_2}=a+1,5b=\dfrac{2,352}{22,4}=0,105\left(mol\right)\) (2)

(1)(2) => a = 0,015 (mol); b = 0,06 (mol)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,015.24}{1,98}.100\%=18,18\%\\\%m_{Al}=\dfrac{0,06.27}{1,98}.100\%=81,82\%\end{matrix}\right.\)

b)

\(\left\{{}\begin{matrix}m_{MgCl_2}=0,015.95=1,425\left(g\right)\\m_{AlCl_3}=0,06.133,5=8,01\left(g\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}C_{M\left(MgCl_2\right)}=\dfrac{0,015}{0,16}=0,09375M\\C_{M\left(AlCl_3\right)}=\dfrac{0,06}{0,16}=0,375M\\C_{M\left(HCl.dư\right)}=\dfrac{0,16.1,5-0,015.20,06.2}{0,16}=0,5625M\end{matrix}\right.\)