gọi cthh có dạng KxOy (x,y∈N*) 
ta có : I . x= y.II 
→\(\dfrac{x}{y}=\dfrac{2}{1}\)(tm) 
=> CTHH : K₂O 
=> CTHH theo đề bài là đúng 

\(1.Cu\left(OH\right)_2;NaOH;Ba\left(OH\right)_2\\ pthh:\\ Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\\ NaOH+HCl\rightarrow NaCl+H_2O\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ 2.Cu\left(OH\right)_2\\ pthh:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\\ 3.NaOH;Ba\left(OH\right)_2\\ pthh:\\ Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\\ 2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ 4.Ba\left(OH\right)_2;NaOH\)

\(1.H_2SO_4+2NaOH\rightarrow Na_2SO_{\text{ 4}}+2H_2O\\ 2.CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ 3.SO_3+H_2O\rightarrow H_2SO_4\\ 4.2HNO_3+CuO\rightarrow Cu\left(NO_3\right)_2+H_2O\\ 5.Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\\ 6.Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\\ MgO+HCl\rightarrow MgCl_2+H_2O\\ 8.SO_2+CaO\rightarrow CaSO_3\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
           0,15<----------------------0,15 
\(m_{Fe}=0,15.56=8,4\left(g\right)\\ \Rightarrow\%m_{Fe}=\dfrac{8,4}{10,5}.100\%=80\%\\ \Rightarrow\%m_{Cu}=100\%-80\%=20\%\)

A 
\(PTHH:BaCl_2+H_2SO_{\text{ 4(lo\text{ãng})}}\rightarrow BaSO_4\downarrow+2HCl\\ Na_2CO_3+H_2SO_{4\left(lo\text{ãng}\right)}\rightarrow Na_2SO_{\text{ 4}}+CO_2+H_2O\)

\(n_{Fe_3O_{\text{ 4 }}}=\dfrac{34,8}{232}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
       0,45<-0,3<----------0,15
\(m_{Fe}=0,45.56=25,2\left(g\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\) 
           0,6<--------------------------------------0,3 
\(m_{KMnO_4}=0,6.158=94,8\left(g\right)\)

\(n_S=\dfrac{48}{32}=1,5\left(mol\right)\\ pthh:S+O_2\underrightarrow{t^o}SO_2\) 
          1,5-->1,5
\(V_{O_2}=1,5.22,4=33,6\left(l\right)\\ V_{KK}=V_{O_2}:\dfrac{1}{5}=33,6.\dfrac{1}{5}=168\left(l\right)\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ pthh:2Zn+O_2\underrightarrow{t^o}2ZnO\) 
           0,3----->0,15--->0,3 
\(m_{ZnO}=0,3.81=24,3\left(g\right)\\ V_{O_2}=0,15.22,4=3,36\left(l\right)\)
\(c,PTHH:2KClO_3\xrightarrow[xtMnO_2]{t^o}2KCl+3O_2\uparrow\) 
                    0,1<------------------------------0,15 
\(m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

theo đề bài ta có : 
p+e-n = 12 
⇔ 2p-n=12 
⇔ n = 2p -12(1) 
Và 
2p + n = 40  (2) 
thay (1) vào (2) ta có 
2p + 2p - 12 = 40 
⇔ 4p = 52 
⇔ p= 13 
⇒ e = p = 13 ; n = 2p-13 = 2.13-12 = 14