\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư

\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

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

b)nAl = 2.7/27=0.1mol

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

(mol) 0.1 0.3 0.1 0.15

VH2 = 0.15*22.4=3.36l

mAlCl3= 0.1*133.5=13.35g

=>C%=\(\dfrac{13,35}{500+27}100=2,533\%\)

a) \(n_{Al}=\dfrac{8,64}{27}=0,32\left(mol\right)\)

\(n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ \(\dfrac{0,32}{2}< \dfrac{1}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

         0,32-->0,96---->0,32--->0,48

=> \(V_{H_2}=0,48.22,4=10,752\left(l\right)\)

b) Trong Y chứa AlCl₃ và HCl dư

\(m_{AlCl_3}=0,32.133,5=42,72\left(g\right)\)

c) m_{dd sau pư} = 8,64 + 365 - 0,48.2 = 372,68 (g)

 \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{42,72}{372,68}.100\%=11,463\%\\C\%\left(HCldư\right)=\dfrac{\left(1-0,96\right).36,5}{372,68}.100\%=0,392\%\end{matrix}\right.\)

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

                   a_____2a______a_____a      (mol)

                \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                    b_____3b_______b_____\(\dfrac{3}{2}\)b         (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27b=36,1\\a+\dfrac{3}{2}b=\dfrac{21,28}{22,4}=0,95\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,5\cdot56=28\left(g\right)\\m_{Al}=8,1\left(g\right)\end{matrix}\right.\)

b+c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl}=2a+3b=1,9\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{1,9}{0,2}=9,5\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư 

\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)

b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\) 

Câu b) là tính nồng độ trong X chứ :v

a) 

PTHH: Mg + 2HCl --> MgCl₂ + H₂

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

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

=> 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

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`0,2`   `0,6`               `0,2`        `0,3`        `(mol)`

`n_[Al] = [ 5,4 ] / 27 = 0,2 (mol)`

`b) V_[H_2] = 0,3 . 22,4 = 6,72 (l)`

`c) V_[dd HCl] = [ 0,6 ] / [ 2,4 ] = 0,25 (l)`

 `=> C_[M_[AlCl_3]] = [ 0,2 ] / [ 0,25 ] = 0,8 (M)`

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

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

0,2        0,6          0,2                0,3   ( mol )

\(V_{H_2}=0,3.22,4=6,72l\)

\(V_{HCl}=\dfrac{0,6}{2,4}=0,25l\)

\(C_{M_{AlCl_3}}=\dfrac{0,2}{0,25}=0,8M\)

2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g