mik lm rồi mà

a)
$\rm (1)2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$
$\rm (2)AlCl_3 + 3NaOh \rightarrow Al(OH)_3 \downarrow + 3NaCl$
$\rm (3)2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O$
$\rm (4)2Al_2O_3 \xrightarrow[Criolit]{t^o} 4Al + 3O_2 \uparrow$
b)

$\rm (1)FeO + H_2 \xrightarrow{t^o} Fe + H_2O$
$\rm (2)Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow$
$\rm (3)FeCl_2 + Zn \rightarrow ZnCl_2 + Fe \downarrow$
$\rm (4)2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
c)

$\rm (1)Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
$\rm (2)2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$\rm (3)FeCl_3 + Al_{dư} \rightarrow AlCl_3 + Fe \downarrow$
$\rm (4)Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow$
d) 

$\rm (1)MgO + 2HCl \rightarrow MgCl_2 + H_2O$
$\rm (2)MgCl_2 + 2KOH \rightarrow Mg(OH)_2 \downarrow + 2KCl$
$\rm (3)Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$
$\rm (4)MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O$
e)

$\rm (1)Zn + 2HCl \rightarrow ZnCl_2 + H_2 \uparrow$
$\rm (2)ZnCl_2 + 2KOH \rightarrow Zn(OH)_2 \downarrow + 2KCl$
$\rm (3)Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O$
$\rm (4) ZnSO_4 + BaCl_2 \rightarrow BaSO_4 \downarrow + ZnCl_2$
f)

$\rm (1)CaO + H_2O \rightarrow Ca(OH)_2$
$\rm (2)Ca(OH)_2 + Na_2CO_3 \rightarrow CaCO_3 \downarrow + 2NaOH$
$\rm (3)CaCO_3 \xrightarrow{t^o} CaO + CO_2 \uparrow$
$\rm (4)CaO + 2HCl \rightarrow CaCl_2 + H_2O$

1:3:2:3

Tỉ lệ \(=\) \(1\) \(:\) \(3\)  \(:\) \(2\) \(:\) \(3\)

a) Gọi CTHH của hợp chất là Mg_xO_y

=> \(\dfrac{24x}{16y}=\dfrac{60}{40}\Leftrightarrow\dfrac{x}{y}=\dfrac{1}{1}\)

=> CTHH có dạng (MgO)_n

Mà PTK = 40 (đvC)

=> \(n=\dfrac{40}{40}=1\)

=> CTHH là MgO

b) Sửa: 25,54%S => 22,54%S thì cộng vào mới đủ 100% :v

1 mol chất có \(\left\{{}\begin{matrix}n_{Na}=\dfrac{32,39\%.142}{23}=2\left(mol\right)\\n_S=\dfrac{22,54\%.142}{32}=1\left(mol\right)\\n_O=\dfrac{142-2.23-32}{16}=4\left(mol\right)\end{matrix}\right.\)

=> CTHH của Na₂SO₄

K.L.P.T= 27.2+16.3=102

\(a,\%Al=27.2.100\div102=52,94\%\)

\(\%O=100-52,94=47,06\%\)

K.L.P.T=39.2+12+16.3=138

\(\%K=39.2.100\div138=56,52\%\)

\(\%C=12.100\div138=8,70\%\)

\(\%O=100-\left(56,52+8,7\right)=34,78\%\)

\(c,\) K.L.P.T= 56.3+16.4=232

\(\%Fe=56.3.100\div232=72,41\%\)

\(\%O=100-72,41=27,59\)

\(d,\) K.L.P.T= 24+12+16.3=84 <amu>

\(\%Mg=24.100\div84=28,57\%\)

\(\%C=12.100\div84=14,29\%\)

\(\%O=100-28,57-14,29=57,14\%\)

$\rm - P_2O_5:$

$\rm 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$\rm P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$
$\rm - Cu(OH)_2$
$\rm CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O$
$\rm CuSO_4 + 2NaOH \rightarrow Cu(OH)_2 \downarrow + Na_2SO_4$
$\rm - Cu(NO_3)_2$
$\rm CuSO_4 + Ba(NO_3)_2 \rightarrow BaSO_4 \downarrow + Cu(NO_3)_2$
$\rm - Na_3PO_4$
$\rm 3NaOH + H_3PO_4 \rightarrow Na_3PO_4 + 3H_2O$

_______________________________________________

$\rm a) Na_2O + H_2O \rightarrow 2NaOH$
$\rm b) BaO + H_2O \rightarrow Ba(OH)_2$
$\rm d) CuSO_4 + 2NaOH \rightarrow Cu(OH)_2 \downarrow + 2NaCl$
$\rm e) FeCl_2 + Ba(OH)_2 \rightarrow BaCl_2 + Fe(OH)_2 \downarrow$