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

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo pt: \(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow C_MHCl=\dfrac{0,2}{0,5}=0,4M\)

Fe + 2HCl → FeCl₂ + H₂

n_Fe = \(\dfrac{5,6}{56}\) = 0,1 (mol)

Theo PT:n_HCl = 2n_Fe = 0,2 (mol)

⇒ C_MHCl = \(\dfrac{0,2}{0,5}\) = 0,4M

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

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

\(a)C+O_2\xrightarrow[]{t^0}CO_2\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\\ b)FeSO_4+BaCl_2\rightarrow FeCl_2+BaSO_4\\ FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\\ Fe\left(OH\right)_2\xrightarrow[]{t^0}FeO+H_2\\ c)2Mg+O_2\xrightarrow[]{t^0}2MgO\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ MgSO_4+BaCl_2\rightarrow MgCl_2+BaSO_4\)

\(a)n_{Zn}=\dfrac{9,1}{65}=0,14mol\\ Zn+CuSO_4\rightarrow Cu+ZnSO_4\\ n_{Cu}=n_{CuSO_4}=n_{Zn}=0,14mol\\ m_{Cu}=0,14.64=8,96g\\ b)m_{CuSO_4}=0,14.160=22,4g\\ m_{ddCuSO_4}=\dfrac{22,4}{20\%}\cdot100\%=112g\\ V_{ddCuSO_4}=\dfrac{112}{1,12}=100ml\)