\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1       0,2          0,1           0,1
\(n_{Fe_2O_3}=\dfrac{21,6-56.0,1}{160}=0,1mol\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\)
0,1            0,6            0,2              0,3
\(V_{ddHCl}=\dfrac{0,2+0,6}{1}=0,8l\\ b.C_{M_{FeCl_2}}=\dfrac{0,1}{0,8}=0,125M\\ C_{M_{FeCl_3}}=\dfrac{0,2}{0,8}=0,25M\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\\ 2NaOH+Zn\rightarrow Na_2ZnO_2+H_2\\ Zn+4HNO_3\rightarrow Zn\left(NO_3\right)_2+2NO_2+2H_2O\)

Ta có: \(n_{Cl_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

____0,2_______0,4______0,2______0,2 (mol)

\(\Rightarrow V_{NaOH}=\dfrac{0,4}{4}=0,1\left(l\right)\)

\(C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

a)

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

b)

\(m_{Mg}=n_{Mg}.24=n_{H_2}.24=0,5.24=12\left(g\right)\Rightarrow m_{Cu}=20-12=8\left(g\right)\)

c)

\(n_{HCl}=2n_{H_2}=2.0,5=1\left(mol\right)\\ V_{HCl}=\dfrac{1}{CM_{HCl}}\) thiếu dữ kiện CM HCl nhe

Ta có: \(n_{Cl_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(2A+nCl_2\rightarrow2ACl_n\)

Theo PT: \(n_A=\dfrac{2}{n}n_{Cl_2}=\dfrac{0,2}{n}\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{4,6}{\dfrac{0,2}{n}}=23n\left(g/mol\right)\)

Với n = 1 thì M_A = 23 (g/mol)

→ A là Na.

Help =((

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,4.56}{35}.100\%=64\%\\\%m_{Cu}=36\%\end{matrix}\right.\)

giúp mình với =((

Câu 1

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Câu 2

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\\ Fe_2O_3+H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)