`1.NaOH+HCl->NaCl+H_2O`

`2.K_2O+H_2O->2KOH`

`3.CuO+H_2SO_4->CuSO_4+H_2O`

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

`5.CaO+CO_2->CaCO_3`

`6.SO_2+Ca(OH)_2->CaSO_3+H_2O`

`7.BaCl_2+H_2SO_4->BaSO_4+2HCl`

`8.KOH+CuCl_2->Cu(OH)_2+2KCl`

`9.2Fe(OH)_3->(t^o)Fe_2O_3+3H_2O`

`10.Na_2CO_3+CaCl_2->CaCO_3+2NaCl`

\(\dfrac{n_{C_xH_y}}{n_{O_2}}=\dfrac{1}{10}\) \(\Rightarrow\left\{{}\begin{matrix}n_{C_xH_y}=1\\n_{O_2}=10\end{matrix}\right.\) ( mol )

\(C_xH_y+\left(x+\dfrac{y}{4}\right)O_2\rightarrow xCO_2+\dfrac{y}{2}H_2O\)

   1              \(x+\dfrac{y}{4}\)             \(x\)            \(\dfrac{y}{2}\)          ( mol )

Cho Y qua dd H₂SO₄ đặc `=>` H₂O bị hấp thụ 

hh khí Z gồm CO₂ và O₂ dư

\(n_{O_2\left(dư\right)}=10-x-\dfrac{y}{4}\left(mol\right)\)

\(d_{\dfrac{hhk}{H_2}}=19\)

\(\Rightarrow M_{hhk}=38\) \((g/mol)\)

Áp dụng quy tắc đường chéo, ta có:

\(\dfrac{n_{CO_2}}{n_{O_2\left(dư\right)}}=\dfrac{1}{1}\) \(\Rightarrow n_{CO_2}=n_{O_2\left(dư\right)}\)

\(\Leftrightarrow x=10-x-\dfrac{y}{4}\) \(\Leftrightarrow2x+\dfrac{y}{4}=10\) \(\Leftrightarrow8x+y=40\)

Với \(x=4;y=8\) thỏa mãn

`=>` CTPT X: \(C_4H_8\)

Đặt \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\\n_{CaCO_3}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_{hh}=106x+100y=30,6\left(g\right)\)  \(\left(1\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

      x                 x                                x                     ( mol )

\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2\uparrow+H_2O\)

     y               y                              y                       ( mol )

\(n_{CO_2}=x+y=0,3\left(mol\right)\)   \(\left(2\right)\)

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

\(\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{30,6}.100=34,64\%\\\%m_{CaCO_3}=100\%-34,64\%=65,36\%\end{matrix}\right.\)

\(C\%_{H_2SO_4}=\dfrac{0,3.98}{225}.100=13,06\%\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)  ( p.ứ thế )

0,3        0,45                0,15       0,45    ( mol )

\(m_{Al}=0,3.27=8,1\left(g\right)\)

\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)

\(C1.m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(C2.BTKL:m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Leftrightarrow8,1+44,1=m_{Al_2\left(SO_4\right)_3}+0,45.2\)

\(\Leftrightarrow m_{Al_2\left(SO_4\right)_3}=51,3\left(g\right)\)

\(n_{ZnCl_2}=\dfrac{100.10\%}{136}=\dfrac{5}{68}\left(mol\right)\)
\(Mg+ZnCl_2\rightarrow MgCl_2+Zn\)

\(\dfrac{5}{68}\)      \(\dfrac{5}{68}\)                                  ( mol )

\(m_{Mg}=\dfrac{5}{68}.24=1,76\left(g\right)\)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,76}{100}.100=1,76\%\\\%m_{Cu}=100\%-1,76\%=98,24\%\end{matrix}\right.\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

\(Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\)

\(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)

\(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

\(2FeCl_3+Fe\rightarrow\left(t^o\right)3FeCl_2\)

\(FeCl_2+KOH\rightarrow Fe\left(OH\right)_2\downarrow+2KCl\)

\(Fe\left(OH\right)_2\xrightarrow[chân.không]{t^o}FeO+H_2O\)

\(FeO+H_2\rightarrow\left(t^o\right)Fe+H_2O\)

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

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

\(H_2O+SO_3\rightarrow H_2SO_4\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(K_2SO_4+BaCl_2\rightarrow2KCl+BaSO_4\downarrow\)

\(2KCl\rightarrow\left(đpnc\right)2K+Cl_2\uparrow\)

\(2Na+Cl_2\rightarrow\left(t^o\right)2NaCl\)

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

\(n_{NaOH}=\dfrac{20.20\%}{40}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{16.65\%}{98}=0,106\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

   0,1      <    0,106                                ( mol )

    0,1            0,05            0,05              ( mol )

\(m_{H_2SO_4\left(dư\right)}=\left(0,106-0,05\right).98=5,5\left(g\right)\)

\(m_{ddspứ}=20+16=36\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{0,05.144}{36}.100=20\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{5,5}{36}.100=15,27\%\end{matrix}\right.\)

Cho hỗn hợp khí đi qua dd Ca(OH)₂ dư, Cl₂ và CO₂ bị hấp thụ, khí O₂ bay lên tiếp tục dẫn qua dd H₂SO₄ đặc để hấp thụ nước `->` Ta thu được O₂ tinh khiết

PTHH:

\(2Ca\left(OH\right)_2+2Cl_2\rightarrow CaCl_2+Ca\left(OCl\right)_2+2H_2O\)

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

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

0,15       0,15                      0,15    ( mol )

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(m_{hh}=\dfrac{8,4}{\left(100-60\right)\%}=21\left(g\right)\)

\(C_{M\left(H_2SO_4\right)}=\dfrac{0,15}{0,05}=3\left(M\right)\)