a+b) PTHH: \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

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

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cl_2}=0,3\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=0,3\cdot71=21,3\left(g\right)\\V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\end{matrix}\right.\) 

PTHH: \(S+O_2\xrightarrow[]{t^o}SO_2\)

Theo PTHH: \(V_{O_2}=22,4\cdot20\%=4,48\left(l\right)=V_{SO_2}\) 

\(\Rightarrow n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_S\) \(\Rightarrow m_S=0,2\cdot32=6,4\left(g\right)\)

PTHH: \(2Mg+O_2\xrightarrow[]{t^o}2MgO\)

Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)=n_{MgO}\) \(\Rightarrow m_{MgO}=0,3\cdot40=12\left(g\right)\)

11) \(Cu\left(NO_3\right)_2+2NaOH\rightarrow2NaNO_3+Cu\left(OH\right)_2\)

12) \(4Fe\left(OH\right)_2+O_2+2H_2O\xrightarrow[]{t^o}4Fe\left(OH\right)_3\)

13) \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

14) \(KNO_3\xrightarrow[]{t^o}KNO_2+\dfrac{1}{2}O_2\)

15) \(C_4H_{10}+\dfrac{13}{2}O_2\xrightarrow[]{t^o}4CO_2+5H_2O\)

16) \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)

17) \(6Na+2H_3PO_4\rightarrow2Na_3PO_4+3H_2\)

18) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

19) \(4NO_2+O_2+2H_2O\rightarrow4HNO_3\)

20) \(Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\)

1) \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

2) \(N_2O_5+H_2O\rightarrow2HNO_3\)

3) \(SO_2+\dfrac{1}{2}O_2\xrightarrow[V_2O_5]{t^o}SO_3\)

4) \(C_2H_2+\dfrac{5}{2}O_2\xrightarrow[]{t^o}2CO_2+H_2O\)

5) \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

6) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

7) \(Al_2\left(SO_4\right)_3+3Ba\left(NO_3\right)_2\rightarrow3BaSO_4+2Al\left(NO_3\right)_3\)

8) \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\)

9) \(Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3+3KNO_3\)

10) \(4FeS_2+11O_2\xrightarrow[]{t^o}2Fe_2O_3+8SO_2\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot9,8\%}{98}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\) \(\Rightarrow\) Fe₂O₃ còn dư, tính theo axit 

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,1\cdot400=40\left(g\right)\\m_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\cdot160\approx5,3\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{40}{16+200-5,3}\cdot100\%\approx18,98\%\)

PT chữ: Natri + Oxi \(--t^o->\) Natri oxit 

PTHH: \(4Na+O_2\xrightarrow[]{t^o}2Na_2O\)

Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\) \(\Rightarrow n_{O_2}=\dfrac{0,1}{4}=0,025\left(mol\right)\) \(\Rightarrow m_{O_2}=0,025\cdot32=0,8\left(g\right)\)

Câu 1:

AlCl₃, Al₂(SO₃)₃, Al₂(SO₄)₃, AlPO₄ và Al(OH)₃

Câu 2: 

Al(OH)₃

Ca(OH)₂

NaOH

Fe(OH)₃

*Sửa đề: "35 gam CaO"

PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)

a) Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{120}{100}=1,2\left(mol\right)=n_{CaO\left(lý.thuyết\right)}\\n_{CaO\left(thực\right)}=\dfrac{35}{56}=0,625\left(mol\right)\end{matrix}\right.\) \(\Rightarrow H\%=\dfrac{0,625}{1,2}\cdot100\%\approx52,08\%\)

b) Với hiệu suất 75% \(\Rightarrow n_{CaO\left(thực\right)}=1,2\cdot75\%=0,9\left(mol\right)\) \(\Rightarrow m_{CaO}=0,9\cdot56=50,4\left(g\right)\)

a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)

c) \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

d) \(Cu+Cl_2\xrightarrow[]{t^o}CuCl_2\)

e) \(2K+S\xrightarrow[]{t^o}K_2S\)