TN1:

PTHH: FeO + H₂ --t^o--> Fe + H₂O

            Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

            CuO + H₂ --t^o--> Cu + H₂O

=> \(n_{O\left(oxit\right)}=n_{H_2O}=\dfrac{15,3}{18}=0,85\left(mol\right)\)

TN2:

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

             Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

             CuO + 2HCl --> CuCl₂ + H₂O

=> \(n_{H_2O}=n_{O\left(oxit\right)}=0,85\left(mol\right)\)

=> n_HCl = 1,7 (mol)

Theo ĐLBTKL: m_oxit + m_HCl = m_muối + m_H2O

=> 50,8 + 1,7.36,5 = m_muối + 0,85.18

=> m_muối = 97,55 (g)

Thiếu đề thì phải

giải giúp mình câu a đk ạ?
Mình cần gấp

\(\left\{{}\begin{matrix}nCu=3x\\nFe3O4=x\end{matrix}\right.\)=>64.3x+232x=42,4

=>x=1

->n Cu=0,3, n Fe3O4=0,1 mol

Fe3O4+8HCl->2FeCl3+FeCl2+4H2O

0,1----------------------0,2

Cu+FeCl3->CuCl2+FeCl3

0,1        0,2

=>n Cu du2=0,2 mol

->m =12,8g

->C

B

Quy đổi Fe₃O₄ thành FeO, Fe₂O₃

\(n_{FeCl_2}=\dfrac{7,62}{127}=0,06\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

            0,06<------------0,06

=> \(n_{Fe_2O_3}=\dfrac{9,12-0,06.72}{160}=0,03\left(mol\right)\)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

             0,03-------------->0,06

=> \(m_{FeCl_3}=0,06.162,5=9,75\left(g\right)\)

CuO+H2->Cu+H2O(1)

Fe2O3+3H2->2Fe+3H2O(2)

Fe3O4+4H2->3Fe+4H2O(3)

nH2O=0.8(mol)

Theo pthh(1)(2)(3) nH2O=nH2

->nH2 cần dùng=0.8(mol)

->V=0.8*22.4=17.92(l)

mH2=0.8*2=1.6(g)

Theo đlbtkl:mOxit+mH2=m nước+m kim loại

<->47.2+1.6=14.4+m kim loại

->m kim loại=47.2+1.6-14.4=34.4(g)

Cảm ơn bạn nhé

Bài 1:

a) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+2H_2O\)

                \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20\cdot80\%=16\left(g\right)\\m_{CuO}=20-16=4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,35\left(mol\right)\) \(\Rightarrow V_{H_2}=0,35\cdot22,4=7,84\left(l\right)\)

c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu}=n_{CuO}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{hhB}=m_{Fe}+m_{Cu}=0,2\cdot56+0,05\cdot64=14,4\left(g\right)\)

Bài 2:

PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

            \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

a) Vì khối lượng Cu bằng \(\dfrac{6}{5}\) khối lượng Fe 

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=\dfrac{26,4}{6+5}\cdot6=14,4\left(g\right)\\m_{Fe}=26,4-14,4=12\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\\n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\) \(\Rightarrow V_{H_2}=\dfrac{153}{280}\cdot22,4=12,24\left(l\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{28}\left(mol\right)\\n_{CuO}=n_{Cu}=0,225\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=\dfrac{3}{28}\cdot160\approx17,14\left(g\right)\\m_{CuO}=0,225\cdot80=18\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=35,14\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{17,14}{35,14}\cdot100\%\approx48,78\%\\\%m_{CuO}=51,22\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

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

\(n_{HCl}=2n_{H_2}=0.5\cdot2=1\left(mol\right)\)

\(BTKL:\)

\(m_X+m_{HCl}=m_M+m_{H_2}\)

\(\Rightarrow m_M=13.4+1\cdot36.5-0.5\cdot2=48.94\left(g\right)\)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe +2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{Cl} = n_{HCl} = 2n_{H_2} = 2.\dfrac{11,2}{22,4} = 1(mol)\\ m_{muối} = m_{kim\ loại} + m_{Cl} = 13,4 + 1.35,5 = 48,9(gam)\)