Bài 4:

A: CO

B: H₂

X: Al

D: Zn

M: S

N: HCl

Q: Cl₂

P: H₂SO₄

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(3Fe_3O_4+8Al\underrightarrow{t^o}4Al_2O_3+9Fe\)

\(FeSO_4+Zn\rightarrow ZnSO_4+Fe\)

\(Fe+S\underrightarrow{t^o}FeS\)

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

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Bài 5:

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(MgCl_2\underrightarrow{đpnc}Mg+Cl_2\)

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+Na_2SO_4\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

1) \(Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\)

2) \(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

3) \(Fe_xO_y+yH_2SO_4\rightarrow Fe_x\left(SO_4\right)_y+yH_2O\)

4) \(2M+2nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nSO_2+2nH_2O\)

5) \(3M+4nHNO_3\rightarrow3M\left(NO_3\right)_n+nNO+2nH_2O\)

6) \(2Fe_xO_y+2yH_2SO_4\rightarrow xFe_2\left(SO_4\right)_{\dfrac{2y}{x}}+0SO_2+2yH_2O\)

7) \(\left(5x-2y\right)Fe_3O_4+\left(46x-18y\right)HNO_3\rightarrow\left(15x-6y\right)Fe\left(NO_3\right)_3+N_xO_y+\left(23x-9y\right)H_2O\)

1) \(C_nH_{2n}+\dfrac{3n}{2}O_2\underrightarrow{t^o}nCO_2+nH_2O\)

2) \(C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\underrightarrow{t^o}nCO_2+\left(n+1\right)H_2O\)

3) \(C_nH_{2n-2}+\dfrac{3n-1}{2}O_2\underrightarrow{t^o}nCO_2+\left(n-1\right)H_2O\)

4) \(C_nH_{2n-6}+\dfrac{3n-3}{2}O_2\underrightarrow{t^o}nCO_2+\left(n-3\right)H_2O\)

5) \(C_nH_{2n+2}O+\dfrac{3n}{2}O_2\underrightarrow{t^o}nCO_2+\left(n+1\right)H_2O\)

1) \(n_{Ba\left(OH\right)_2}=0,2.0,1=0,03\left(mol\right)\)

\(n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)\)

TH1: Nếu kết tủa không bị hòa tan 

PTHH: Ba(OH)₂ + CO₂ --> BaCO₃ + H₂O

                              0,02<---0,02

=> n_CO2 = 0,02 (mol)

\(n_{O_2\left(pư\right)}=\dfrac{8+0,02.44-9,28}{32}=-0,0125\)  (vô lí)

TH2: Nếu kết tủa bị hòa tan một phần

PTHH: Ba(OH)₂ + CO₂ --> BaCO₃ + H₂O

              0,03---->0,03----->0,03

            BaCO₃ + CO₂ + H₂O ---> Ba(HCO₃)₂

             0,01---->0,01

=> n_CO2 = 0,04 (mol)

=> \(n_{FeCO_3}=0,04\left(mol\right)\)

\(n_{O_2\left(pư\right)}=\dfrac{8+0,04.44-9,28}{32}=0,015\left(mol\right)\)

Gọi số mol Fe_xO_y trong A là a (mol)

Oxit sau pư là Fe₂O₃

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

Bảo toàn Fe: 0,04 + ax = 0,05.2

=> ax = 0,06 (mol)

Bảo toàn O: 0,04.3 + ay + 0,015.2 = 0,05.3 + 0,04.2

=> ay = 0,08 (mol)

Xét \(\dfrac{x}{y}=\dfrac{ax}{ay}=\dfrac{0,06}{0,08}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) \(n_{Zn}=\dfrac{16}{65}\left(mol\right)\)

Theo PTHH: \(n_{H_2}=n_{Zn}=\dfrac{16}{65}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{16}{65}.22,4=\dfrac{1792}{325}\left(l\right)\)

c) Theo PTHH: \(n_{HCl}=2.n_{Zn}=\dfrac{32}{65}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{32}{65}.36,5=\dfrac{1168}{65}\left(g\right)\)

a) \(\left\{{}\begin{matrix}56.n_{Fe}+27.n_{Al}=16,4\\n_{Fe}:n_{Al}=1:4\end{matrix}\right.\) => \(\left\{{}\begin{matrix}n_{Fe}=0,1\left(mol\right)\\n_{Al}=0,4\left(mol\right)\end{matrix}\right.\)

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

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

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

            0,4--->1,2------->0,4--->0,6

\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68\left(l\right)\)

n_HCl(pư) = 0,2 + 1,2 = 1,4 (mol)

=> \(n_{HCl\left(TT\right)}=\dfrac{1,4.110}{100}=1,54\left(mol\right)\)

=> \(x\%=\dfrac{1,54.36,5}{500}.100\%=11,242\%\)

b) m_{dd sau pư} = 16,4 + 500 - 0,7.2 = 515 (g)

\(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{515}.100\%=2,466\%\\C\%_{AlCl_3}=\dfrac{0,4.133,5}{515}.100\%=10,369\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(1,54-1,4\right).36,5}{515}.100\%=0,992\%\end{matrix}\right.\)

c) \(V_{dd.A}=\dfrac{515}{1,25}=412\left(ml\right)=0,412\left(l\right)\)

\(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{1,54-1,4}{0,412}=0,34M\\C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,412}=0,243M\\C_{M\left(AlCl_3\right)}=\dfrac{0,4}{0,412}=0,971M\end{matrix}\right.\)

\(n_{NaOH}=\dfrac{400.2\%}{40}=0,2\left(mol\right)\)

PTHH: \(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

               0,2------>0,2

=> \(C_{M\left(dd.HNO_3\right)}=\dfrac{0,2}{0,2}=1M\)

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

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PTHH: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

=> \(\%m_{MgO}=\dfrac{12-0,1.24}{12}.100\%=80\%\)

\(n_{MgO}=\dfrac{12-0,1.24}{40}=0,24\left(mol\right)\)

Theo PTHH: \(n_{HCl}=2.n_{Mg}+2.n_{MgO}=0,68\left(mol\right)\)

=> \(m_{dd.HCl}=\dfrac{0,68.36,5}{20\%}=124,1\left(g\right)\)