a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\); \(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,8}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2-->0,6---->0,2----->0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b) \(\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{HCl\left(dư\right)}=\left(0,8-0,6\right).36,5=7,3\left(g\right)\end{matrix}\right.\)

=> m_{chất tan} = 26,7 + 7,3 = 34 (g)

c) m_{dd sau pư} = 5,4 + 200 - 0,3.2 = 204,8 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{26,7}{204,8}.100\%=13,04\%\\C\%_{HCl\left(dư\right)}=\dfrac{7,3}{204,8}.100\%=3,56\%\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2                       0,2       0,3 
\(V_{H_2}=0,3.22,4=6,72L\\ m_{AlCl_3}=133,5.0,2=26,7g\\ m_{\text{dd}}=5,4+200-\left(0,3.2\right)=204,8g\\ C\%=\dfrac{26,7}{204,8}.100\%=13\%\)

a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

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

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,8}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

c, \(n_{HCl\left(pư\right)}=2n_{Zn}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 19,5 + 200 - 0,3.2 = 218,9 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{218,9}.100\%\approx3,33\%\\C\%_{ZnCl_2}=\dfrac{40,8}{218,9}.100\%\approx18,64\%\end{matrix}\right.\)

\(a)n_{Zn}=\dfrac{19,5}{65}=0,3mol\\ n_{HCl}=\dfrac{200.14,6}{100.36,5}=0,8mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,3mol\\ V_{H_2}=0,3.22,4=6,72l\\ b)m_{ZnCl_2}=0,3.136=40,8g\\ c)n_{HCl.pư}=0,3.2=0,6mol\\ C_{\%ZnCl_2}=\dfrac{40,8}{200+19,5-0,3.2}\cdot100=18,64\%\\ C_{\%HCl.dư}=\dfrac{\left(0,8-0,6\right).36,5}{200+19,5-0,3.2}\cdot100=3,33\%\)

\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)

Chọn đáp án B

n_H2 = n_CuO = 0,4  (mol)

BTNT H: n_HCl = 2n_H2 = 0,8 (mol)

BTKL: m_muối = m_Al+Zn + m_Cl-  = 11,9 + 0,8.35,5 = 40,3 (g)

a. \(m_{Al.pứ}=15-9,6=5,4\left(g\right)\)

b. \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4.100\%}{15}=36\%\\\%m_{Cu}=\dfrac{9,6.100\%}{15}=64\%\end{matrix}\right.\)

c. \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.\dfrac{5,4}{27}=0,3\left(mol\right)\Rightarrow V_{H_2\left(đkc\right)}=0,3.24,79=7,437\left(l\right)\)

d. \(\%m_{AlCl_3}=\dfrac{0,2.133,5.100\%}{15+200-9,6}=13\%\)

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)