PT: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Ta có: 72n_FeO + 102n_Al2O3 = 45 (1)

\(n_{HCl}=1.2,2=2,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{FeO}+6n_{Al_2O_3}=2,2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{FeO}=0,2\left(mol\right)\\n_{Al_2O_3}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{FeO}=0,2.72=14,4\left(g\right)\\m_{Al_2O_3}=0,3.102=30,6\left(g\right)\end{matrix}\right.\)

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{FeO}=0,2\left(mol\right)\\n_{AlCl_3}=2n_{Al_2O_3}=0,6\left(mol\right)\end{matrix}\right.\) 

⇒ m _muối = m_FeCl2 + m_AlCl3 = 0,2.127 + 0,6.133,5 = 105,5 (g)

\(n_{HCl}=1.2,2=2,2mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ n_{FeO}=a;n_{Al_2O_3}=b\\ \Rightarrow\left\{{}\begin{matrix}72a+102b=45\\2a+6b=2,2\end{matrix}\right.\\ \Rightarrow a=0,2;b=0,3\\ m_{FeO}=0,2.72=14,4g\\ m_{Al_2O_3}=45-14,4=30,6g\\ n_{FeO}=n_{FeCl_2}=0,2mol\\ n_{Al_2O_3}=0,3.2=0,6mol\\ m_{muối}=0,2.127+0,6.133,5=105,5g\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(n_{HCl}=0.35\cdot1=0.35\left(mol\right)\)

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

Lập tỉ lệ : 

\(0.1< \dfrac{0.35}{2}\Rightarrow HCldư\)

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

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

               2Al   +   6HCl   →   2AlCl₃   +   3H₂

( mol )  0,2                       →    0, 2

\(m_{AlCl_3}=n.M=0,2.\left(27+35,5\times3\right)=26,7\left(g\right)\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

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

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

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

Bảo toàn nguyên tố Fe:

\(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=12,7\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{ZnSO_4}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnSO_4}=0,1.161=16,1(g)\)

Bảo toàn nguyên tố Zn:

\(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=16,1\left(g\right)\)

\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{200\cdot14.6\%}{36.5}=0.8\left(mol\right)\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(TC:\)

\(\dfrac{0.1}{1}< \dfrac{0.8}{6}\Rightarrow HCldư\)

\(m_{HCl\left(dư\right)}=\left(0.8-0.6\right)\cdot36.5=7.3\left(g\right)\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

Gọi số mol Fe, Zn là a,b

=> 56a + 65b = 12,1 (1)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______a---->2a

Zn + 2HCl --> ZnCl₂ + H₂

_b---->2b

=> 2a + 2b = 0,4 (2)

(1)(2) => \(\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

=> \(\%Fe=\dfrac{56.0,1}{12,1}.100\%=46,28\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=12,1(1)\)

\(n_{HCl}=2.0,2=0,4(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,4(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\)

a, Ta có: \(n_{CO}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{CO_2}\)

Theo ĐLBT KL, có: m_hh + m_CO = m_Fe + m_CO2

⇒ m_Fe = 18,2 + 0,3.28 - 0,3.44 = 13,4 (g)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Ca}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

⇒ x + y = 0,2 (1)

PT: \(Ca+2HCl\rightarrow CaCl_2+H_2\)

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

\(CaCl_2+Na_2CO_3\rightarrow CaCO_{3\downarrow}+2NaCl\)

\(MgCl_2+Na_2CO_3\rightarrow MgCO_{3\downarrow}+2NaCl\)

Theo PT: \(\left\{{}\begin{matrix}n_{CaCO_3}=n_{Ca}=x\left(mol\right)\\n_{MgCO_3}=n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

⇒ 100x + 84y = 18,4 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{CaCl_2}=n_{Ca}=0,1\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ a = m_CaCl2 + m_MgCl2 = 0,1.111 + 0,1.95 = 20,6 (g)

Bạn tham khảo nhé!