a)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)

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

bđ       0,3       0,7

pư      0,3       0,6

spư      0         0,1        0,3       0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b)

m_dd = 16,8 + 175 - 0,3.2 = 191,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{191,2}.100\%=1,91\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{175.14,6}{100}}{36,5}=0,7\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
\(LTL:\dfrac{0,3}{1}< \dfrac{0,7}{2}\)  
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\\ m_{\text{dd}}=16,8+175-\left(0,3.2\right)=191,2\left(g\right)\\ n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\\ C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,92\%\)
=> HCl dư 
 

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

a, \(m_{HCl}=200.14,6\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,4       0,8         0,4         0,4

\(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

b, \(m_{Zn}=0,4.65=26\left(g\right)\)

c, m_{dd sau pứ} = 26 + 200 - 0,4.2 = 225,2 (g)

\(C_{M_{ddZnCl_2}}=\dfrac{0,4.136.100\%}{225,2}=24,16\%\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1........2\)

\(0.1......0.4\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)

\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)

\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)

\(n_{Zn}=\frac{13}{65}=0,2mol\)

\(PTHH:\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0,2\rightarrow0,4\rightarrow0,2\rightarrow0,2\left(mol\right)\)

a)\(V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48\left(l\right)\)

b)\(m_{ddspư}=m_{Zn}+m_{ddHCl}-m_{H_2}\)

                \(=13+100-0,2.1=112,8\left(gam\right)\)

\(C\%_{ddspư}=\frac{m_{ZnCl_2}.100}{m_{ddspư}}=\frac{0,2.136.100}{112,8}\approx24,1\left(\%\right)\)

\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)

1.

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

m_HCl=\(\dfrac{175.14,6}{100}\)=25,55g

n_HCl=\(\dfrac{25,55}{36,5}\)=0,7

                     2Al + 6HCl → 2AlCl₃ + 3H₂↑

n trước pứ    0,2     0,7  

n pứ              0,2 →0,6  →    0,2  →  0,3  mol

n sau pứ       hết   dư 0,1

Sau pứ HCl dư.

m_HCl (dư)= 36,5.0,1=3,65g

m_{các chất sau pư}= 5,4 +175 - 0,3.2= 179,8g

m_AlCl3= 133,5.0,2=26,7g

C%_ddHCl (dư)= \(\dfrac{3,65.100}{179,8}=2,03%\)%

C%_ddAlCl3 = \(\dfrac{26,7.100}{179,8}\)= 14,85%

2.

           200ml= 0,2l

m_Mg= \(\dfrac{4,2}{24}=0,175mol\)

    Mg + 2HCl → MgCl₂ + H₂↑

0,175→ 0,35 → 0,175→0,175 mol

a) V_H2= 0,175.22,4=3,92l.

b)C%_dHCl= \(\dfrac{0,35}{0,2}=1,75\)M

`n_[Al]=[2,7]/27=0,1(mol)`

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

`0,1`    `0,3`              `0,1`      `0,15`           `(mol)`

`a)V_[H_2]=0,15.22,4=3,36(l)`

`b)V_[dd HCl]=[0,3]/2=0,15(l)`

`=>C_[M_[AlCl_3]]=[0,1]/[0,15]~~0,67(M)`

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
          0,1        0,3            0,1            0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)