Gọi $n_{Mg} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 24a + 65b = 15,3(1)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)$

Từ (1)(2) suy ra : $a = \dfrac{81}{410} ; b = \dfrac{21}{205}$

$m_{Mg} = \dfrac{81}{410}.24 = 4,74(gam)$
$m_{Zn} = 15,3 - 4,74 = 10,56(gam)$

a)

Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$

$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1; b = 0,2

$m_{Zn} = 0,1.65 = 6,5(gam)$

$m_{Al} = 0,2.27 = 5,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$

Em tham khảo nhé!

Ta có: \(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_S=\dfrac{4,8}{32}=0,15\left(mol\right)\)

Coi hh chất rắn gồm M và O.

⇒ n_O = 0,15.2 = 0,3 (mol)

Ta có: \(n_M=\dfrac{16,2}{M_M}\left(mol\right)\)

BT e, có: n.n_M = 2n_O + 2n_SO2 + 6n_S 

\(\Rightarrow\dfrac{16,2n}{M_M}=1,8\Rightarrow M_M=9n\left(g/mol\right)\)

Với n = 3 thì M_M = 27 (g/mol) là thỏa mãn.

Vậy: M là Al.

a) Nồng độ.

b) Nhiệt độ.

c) Xúc tác.

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\)

\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(m_{H_2O}=0,6.18=10,8\left(g\right)\)

\(m_{muoi.khan}=37,6+43,8-10,8=70,6\left(g\right)\) ( định luật bảo toàn khối lượng )

Gọi n Fe = a (mol )

       n Mg = b (mol )  (a,b > 0)

--> 56a+24b = 13,2 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

a            2a           a           a

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

b             2b             b            b

----> a+b=0,35 

Ta có hệ Pt :

\(\left\{{}\begin{matrix}56a+24b=13,2\\a+b=0,35\end{matrix}\right.\)

Giải hệ PT , ta có : 

a= 0,15 

b = 0,2 (mol )

\(V_{HClđủ}=\left(0,15.2+0,2.2\right):0,5=1,4\left(l\right)\)

\(a,m_{Fe}=0,15.56=8,4\left(g\right)\)

\(m_{Mg}=0,2.24=4,8\left(g\right)\)

\(\%m_{Fe}=\dfrac{8,4}{13,2}.100\%\approx63,64\%\)

\(\%m_{Mg}=\dfrac{4,8}{13,2}.100\%\approx36,36\%\)

\(b,m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(c,HCl+NaOH\rightarrow NaCl+H_2O\)

    0,2        0,2 

\(m_{NaOH}=\dfrac{100.8}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(V_{HCldư}=\dfrac{n}{C_M}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

\(V_{HCl}=V_{HClđủ}+V_{HCldư}=1,4+0,4=1,8\left(l\right)\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,3}{1}>\dfrac{0,5}{2}\Rightarrow HCl.hết,Zn.dư\\ Chất.sau.phản.ứng:ZnCl_2,H_2,Zn\left(dư\right)\\ n_{ZnCl_2}=n_{H_2}=n_{Zn\left(p.ứ\right)}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ n_{Zn\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\\ m_{ZnCl_2}=136.0,25=34\left(g\right)\\ m_{H_2}=0,25.2=0,5\left(g\right)\\ m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\)