$\text{Câu 1:}$

$\text{a. ???}$

$\text{b. $4P_{ }$ + $5O_{2}$ → $2P_{2}O_{5}$}$ 

$\text{c. $2Al_{ }$ + $3CuCl_{2}$ → $2AlCl_{3}$ + $3Cu_{ }↓$}$

$\text{d. $2Mg_{ }$ + $O_{2}$ → $2MgO_{ }$}$

$\text{e. $2Al_{ }$ + $6HCl_{ }$ → $2AlCl_{3}$ + $3H_{2}↑$}$

$\text{f. $FeCl_{3}$ + $3AgNO_{3}$ → $Fe(NO_{3})_{3}$ + $3AgCl_{ }$↓}$

$\text{g. $3Fe_{ }$ + $2O_{2}$ → $Fe_{3}O_{4}$}$

$\text{h. $2Na_{ }$ + $2HCl_{ }$ → $2NaCl_{ }$ + $H_{2}$↑}$

$\text{Câu 2:}$

$\text{a. Ta có: $m_{CuSO_{4}}$ = 64+32+16*4 = 160 g}$

$\text{⇒ %$m_{Cu}$ = 64/160*100% = 40%}$

$\text{⇒ %$m_{S}$ = 32/160*100% = 20%}$

$\text{⇒ %$m_{O}$ = 100%-(40+20)% = 40%}$

$\text{b. Ta có: $m_{H_{3}PO_{4}}$ = 1*3+31+16*4 = 98 g}$

$\text{⇒ %$m_{H}$ = 3/98*100% = 3.06%}$

$\text{⇒ %$m_{P}$ = 31/98*100% = 31.63%}$

$\text{⇒ %$m_{O}$ = 100%-(3.06+31.63)% = 65.31%}$

$\text{Câu 3:}$

$\text{Ta có: $M_{hc}$ = 160 g/mol}$

$\text{⇒ $m_{Cu}$ = %$m_{Cu}$*160 = 40/100*160 = 64 g ⇒ $n_{Cu}$ = 64/64 = 1 mol}$

$\text{⇒ $m_{S}$ = %$m_{S}$*160 = 20/100*160 = 32 g ⇒ $n_{S}$ = 32/32 =1 mol}$

$\text{⇒ $m_{O}$ = 160-(64+32) = 64 g ⇒ $n_{O}$ = 64/16 = 4 mol}$

$\text{Trong 1 mol hợp chất có 1 mol $Cu_{ }$, 1 mol $S_{ }$ và 4 mol $O_{ }$}$

$\text{⇒ CTHH của hc: $CuSO_{4}$}$

$\text{Câu 4:}$

$\text{Ta có: $d_{A/kk}$ = 0.552 ⇒ $M_{A}$ = 0.552*29 = 16 g/mol}$

$\text{$m_{C}$ = 75/100*16 = 12 g ⇒ $n_{C}$ = 12/12 = 1 mol}$

$\text{$m_{H}$ = 16-12 = 4 g ⇒ $n_{H}$ = 4/1 = 4 mol}$

$\text{Trong 1 mol khí A có 1 mol $C_{ }$ và 4 mol $H_{ }$}$

$\text{⇒ CTHH của A: $CH_{4}$}$

$\text{Câu 5:}$

\begin{array}{|c|c|c|c|c|}\hline \text{ }&\text{n(mol)}&\text{m(g)}&\text{$V_{khí}$(lít)(đktc)}&\text{Số phân tử}\\\hline \text{$CO_{2}$}&\text{0.01}&\text{0.44}&\text{0.224}&\text{6*$10^{21}$}\\\hline \text{$N_{2}$}&\text{0.2}&\text{5.6}&\text{4.48}&\text{2*$10^{22}$}\\\hline \text{$SO_{3}$}&\text{0.5}&\text{40}&\text{11.2}&\text{3*$10^{23}$}\\\hline \text{$CH_{4}$}&\text{0.5}&\text{8}&\text{11.2}&\text{3*$10^{23}$}\\\hline\end{array}