\(Fe + 2HCl \to FeCl_2 + H_2\\ n_{Fe} = n_{FeCl_2} = n_{H_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ m_{Fe} = 0,4.56 = 22,4(gam)\\ m_{FeCl_2} = 0,4.127 = 50,8(gam)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

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

\(n_{Fe}=n_{FeCl_2}=n_{H_2}=0.4\left(mol\right)\)

\(m_{Fe}=0.4\cdot56=22.4\left(g\right)\)

\(m_{FeCl_2}=0.4\cdot127=50.8\left(g\right)\)

a) PTHH : Fe + 2HCl → FeCl₂ + H₂

b) V_H2(đktc)=8,96(l) → n_H2= \(\dfrac{8,96}{22,4}\)=0,4(mol)

→ n_Fe= 0,4(mol)

m_Fe= 0,4 . 56=22,4(g)

c) Ta có :  n_H2= \(\dfrac{8,96}{22,4}\)=0,4(mol) → n_FeCl2= 0,4(mol)

m_FeCl2= 0,4 . 127=50,8(g)

Do R thuộc nhóm IIA nên R có hóa trị II.

PT: \(R+2HCl\rightarrow RCl_2+H_2\)

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

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

\(\Rightarrow M_R=\dfrac{9,6}{0,4}=24\left(g/mol\right)\)

Vậy: R là Magie (Mg).

Bạn tham khảo nhé!

8,96l Cl2 + H2 ---H=75%---> 2HCl

0,4............................................0,8

V _{Hcl lí thuyết :} 0,8 . 22,4 = 17,92 (l)

V _{HCl thực tế :} 17,92 . 75% = 13,44 (l)

\(n_{Cl_2}=\dfrac{8,96}{22,4}=0,25\left(mol\right)\)

PT:     Cl₂ + H₂ → 2HCl

Mol:  0,25               0,5

\(m_{HCl\left(lt\right)}=0,5.36,5=18,25\left(g\right)\)

 \(\Rightarrow m_{HCl\left(tt\right)}=75\%.18,25=13,6875\left(g\right)\)

Cl₂+H₂-to>2HCl

0,4------------0,8 mol

n _Cl2=\(\dfrac{8,96}{22,4}\)=0,4 mol

m _{HCl pứ} =\(\dfrac{0,8.36,5.75}{100}\)=21,9kg

a) PTHH : \(Fe+2HCl-t^o->FeCl_2+H_2\)  (1)

                 \(2Fe+3Cl_2-t^o->2FeCl_3\)          (2)

                  \(Cu+Cl_2-t^o->CuCl_2\)              (3)

b) Theo pthh (1) : \(n_{Fe}=n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

Theo pthh (2) và (3) : \(\Sigma n_{Cl2}=\dfrac{3}{2}n_{Fe}+n_{Cu}\)

\(\Rightarrow\dfrac{6,72}{22,4}=\dfrac{3}{2}.0,1+n_{Cu}\)

\(\Rightarrow0,3=0,15+n_{Cu}\)

\(\Rightarrow n_{Cu}=0,15\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{5,6+9,6}\cdot100\%\approx36,84\%\\\%m_{Cu}=100\%-36,84\%=63,16\%\end{matrix}\right.\)

$Fe_2O_3+3H_2\rightarrow 2Fe+3H_2O$

$Fe+2HCl\rightarrow FeCl_2+H_2$

Gọi số mol Fe2O3 là a

Ta có: $n_{H_2/(1)}=3a(mol);n_{H_2/(2)}=2a(mol)$

\(\Rightarrow\dfrac{V}{V'}=\dfrac{n_{H_2\left(1\right)}}{n_{H_2\left(2\right)}}=\dfrac{3}{2}\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{Cl_2}=\dfrac{V_{Cl_2\left(ĐKTC\right)}}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+3Cl_2\rightarrow2AlCl_3\\ \dfrac{n_{Al}}{n_{Cl_2}}=\dfrac{1}{2}< \dfrac{2}{3}\Rightarrow Cl_2\text{ dư, bài toán tính theo }Al\\ \Rightarrow m_{AlCl_3}=n_{AlCl_3}\cdot M_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\)