1/

- Điều chế được các khí: \(SO_2;H_2;O_2;Cl_2;CO_2;H_2S;HCl\)

\(2H_2O\rightarrow\left(đp\right)2H_2+O_2\)

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

\(2NaCl+2H_2O\xrightarrow[đpdd]{cmn}2NaOH+H_2\uparrow+Cl_2\uparrow\)

\(H_2+Cl_2\xrightarrow[as]{t^o}2HCl\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)

\(H_2+S\rightarrow\left(t^o\right)H_2S\)

b. Các oxit axit và oxit bazơ: \(SO_2;CO_2;Fe_3O_4;MgO;Na_2O\)

\(2H_2O\rightarrow\left(đp\right)2H_2+O_2\)

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

\(MgCO_3\rightarrow\left(t^o\right)MgO+CO_2\)

\(2NaCl\rightarrow\left(đpnc\right)2Na+Cl_2\)

\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)

c. Các bazơ: \(NaOH;Mg\left(OH\right)_2;Fe\left(OH\right)_2;Fe\left(OH\right)_3\)

\(2NaCl+2H_2O\xrightarrow[đpdd]{cmn}2NaOH+H_2\uparrow+Cl_2\uparrow\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)

\(NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

\(H_2+Cl_2\xrightarrow[as]{t^o}2HCl\)

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

\(NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)

2/

\(4FeS_2+11O_2\rightarrow\left(t^o\right)2Fe_2O_3+8SO_2\uparrow\)

\(2SO_2+O_2⇌\left(t^o,V_2O_5\right)2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

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

3/

`@`\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2\uparrow+H_2\uparrow\)

`@`\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(Ba\left(OH\right)_2+Na_2CO_3\rightarrow BaCO_3\downarrow+2NaOH\)

4/

`@`\(Ca+Cl_2\rightarrow\left(t^o\right)CaCl_2\)

`@`\(CaSO_4+BaCl_2\rightarrow CaCl_2+BaSO_4\downarrow\)

`@`\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

`@`\(CaS+CuCl_2\rightarrow CaCl_2+CuS\downarrow\)

5/

`@`\(2H_2S+3O_2\rightarrow\left(t^o\right)2SO_2+2H_2O\)

`@`\(4FeS_2+11O_2\rightarrow\left(t^o\right)2Fe_2O_3+8SO_2\uparrow\)

`@`\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow\left(t^o\right)CuSO_4+SO_2\uparrow+2H_2O\)

`@`\(2FeS_2+14H_2SO_{4\left(đ\right)}\rightarrow\left(t^o\right)Fe_2\left(SO_4\right)_3+15SO_2+14H_2O\)

`@`\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2\uparrow+H_2O\)

`@`\(S+O_2\rightarrow\left(t^o\right)SO_2\)

\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

0,35      0,35          0,35      0,35  ( mol )

\(m_{dd_{H_2SO_4}}=\dfrac{0,35.98}{25\%}=137,2\left(g\right)\)

\(m_{ddspứ}=0,35.65+137,2-0,35.2=159,25\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{0,35.161}{159,25}.100=35,38\%\)

1.

\(Fe_xO_y+yCO\rightarrow\left(t^o\right)xFe+yCO_2\)

Giả sử CO khử hết Fe_xO_y

\(n_{Fe}=\dfrac{5,36}{56}=0,095\left(mol\right)\)

\(Fe+4HNO_{3\left(l\right)}\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)

0,095                          0,095                           ( mol )

\(m_{Fe\left(NO_3\right)_3}=0,095.242=22,99\left(g\right)>21,78\left(g\right)\)

\(n_{Fe\left(NO_3\right)_3}=\dfrac{21,78}{242}=0,09\left(mol\right)\)

`->` Chất rắn Y gồm có Fe và Fe_xO_y

Bảo toàn Fe: \(n_{Fe}=n_{Fe\left(NO_3\right)_3}=0,09\left(mol\right)\)

\(n_{O\left(Fe_xO_y\right)}=\dfrac{6,96-\left(0,09.56\right)}{16}=0,12\left(mol\right)\)

\(\dfrac{x}{y}=\dfrac{0,09}{0,12}=\dfrac{3}{4}\)

`=>` CTHH: \(Fe_3O_4\)

Đặt \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Fe_3O_4\left(dư\right)}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_y=56x+232y=5,36\left(1\right)\)

\(Fe+4HNO_{3\left(l\right)}\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)

 x                                    x              x               ( mol )

\(3Fe_3O_4+28HNO_{3\left(l\right)}\rightarrow9Fe\left(NO_3\right)_3+NO+14H_2O\)

  y                                         3y               1/3 y                 ( mol )

\(n_{Fe\left(NO_3\right)_3}=x+3y=0,09\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,075\\y=0,005\end{matrix}\right.\)

\(V_{NO}=\left(0,075+\dfrac{1}{3}.0,005\right).22,4=2,05\left(l\right)\)

2.

Quy hỗn hợp về \(\left\{{}\begin{matrix}Fe=x\\O=y\end{matrix}\right.\) ( mol )

\(m_{hh}=56x+16y=8,96\left(g\right)\left(1\right)\)

\(n_{NO_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

\(Fe^0\rightarrow Fe^{+3}\)

\(N^{+5}\rightarrow N^{+4}\)

\(O^0\rightarrow O^{-2}\)

Bảo toàn e: \(3n_{Fe}=n_{NO}+2n_O\)

\(\Leftrightarrow3x=0,08+2y\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,12\\y=0,14\end{matrix}\right.\)

Bảo toàn Fe: \(n_{Fe}=n_{Fe\left(NO_3\right)_3}=0,12\left(mol\right)\)

\(m_{Fe\left(NO_3\right)_3}=0,12.242=29,04\left(g\right)\)

\(10Al+36HNO_{3\left(l\right)}\rightarrow10Al\left(NO_3\right)_3+3N_2\uparrow+18H_2O\)

\(Al+4HNO_{3\left(l\right)}\rightarrow Al\left(NO_3\right)_3+NO\uparrow+2H_2O\)

\(n_{hhk}=\dfrac{0,1792}{22,4}=8.10^{-3}\left(mol\right)\)

\(\Rightarrow n_{N_2}+n_{NO}=8.10^{-3}\left(1\right)\)

\(d_{\dfrac{X}{H_2}}=14,25\)

\(\Rightarrow M_X=28,5\) \((g/mol)\)

\(\Leftrightarrow\dfrac{28.n_{N_2}+30.n_{NO}}{n_{N_2}+n_{NO}}=28,5\)

\(\Leftrightarrow\dfrac{28.n_{N_2}+30.n_{NO}}{8.10^{-3}}=28,5\)

\(\Leftrightarrow28.n_{N_2}+30.n_{NO}=0,228\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}n_{N_2}=6.10^{-3}\\n_{NO}=2.10^{-3}\end{matrix}\right.\) ( mol )

\(n_{Al}=\dfrac{10}{3}.6.10^{-3}+2.10^{-3}=22.10^{-3}\left(mol\right)\)

\(m_{Al}=22.10^{-3}.27=0,594\left(g\right)\)

Đặt \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_2}=y\end{matrix}\right.\) ( mol )
\(n_{hhk}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) ; \(n_{O_2}=\dfrac{21,28}{22,4}=0,95\left(mol\right)\)

\(\Rightarrow n_{hhk}=x+y=0,4\left(mol\right)\left(1\right)\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

  x           2x               x                  ( mol )

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

   y           2,5y             2y                    ( mol )

\(n_{O_2}=2x+2,5y=0,95\left(mol\right)\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,4}.100=25\%\\\%V_{C_2H_2}=100\%-25\%=75\%\end{matrix}\right.\)

\(\Sigma n_{CO_2}=0,1+2.0,3=0,7\left(mol\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                      0,7           0,7                   ( mol )

\(m_{CaCO_3}=0,7.100=70\left(g\right)\)

a. H.tượng: Mảnh nhôm không tan trong dd

b. H.tượng: nhôm tan trong dd, dd màu xanh nhạt dần, đồng tách ra bám vào mảnh nhôm

\(2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\downarrow\)

c. H.tượng: nhôm tan dần, xuất hiện kết tủa trắng

\(Al+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3Ag\downarrow\)

d.H.tượng: Nhôm tan trong dd, có hiện tượng sủi bọt khí

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

\(X+O_2\rightarrow SO_2+H_2O\)

`=>` S gồm có S,H và có thể có O

Bảo toàn S: \(n_S=n_{SO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

Bảo toàn H: \(n_H=2.n_{H_2O}=2.\dfrac{3,6}{18}=0,4\left(mol\right)\)

\(n_O=\dfrac{6,8-\left(0,2.32+0,4.1\right)}{16}=0\left(mol\right)\)

CTPT \(X:S_xH_y\)

\(\dfrac{x}{y}=\dfrac{0,2}{0,4}=\dfrac{1}{2}\)

`=>` CTPT X: \(SH_2\)

ũa X là \(p+n\) mà