Đặt `n_{N_2}=x(mol);n_{O_2}=y(mol)`

`->x+y={3,36}/{22,4}=0,15(1)`

Có `{28x+32y}/{x+y}={44}/3.2={88}/3`

`->x=2y(2)`

`(1)(2)->x=0,1(mol);y=0,05(mol)`

`n_{N_2}=0,1(mol)->\%V_{N_2}={0,1}/{0,15}.100\%\approx 66,67\%`

`n_{O_2}=0,05(mol)->\%V_{O_2}=100-66,67=33,33\%`

\(\begin{array}{l}
4Na+O_2\xrightarrow{t^o}2Na_2O\\
Na_2O+H_2O\to 2NaOH\\
2NaOH+CO_2\to Na_2CO_3+H_2O\\
Na_2CO_3+BaCl_2\to BaCO_3+2NaCl\\
2NaCl+2H_2O\xrightarrow{đpcmn}2NaOH+Cl_2+H_2\\
NaOH+CO_2\to NaHCO_3\\
2NaHCO_3\xrightarrow{t^o}Na_2CO_3+CO_2+H_2O\\
Na_2CO_3+Ba(OH)_2\to 2NaOH+BaCO_3\\
BaCO_3+2HCl\to BaCl_2+CO_2+H_2O\\
BaCl_2+2AgNO_3\to Ba(NO_3)_2+2AgCl\\
Ba(NO_3)_2+CuSO_4\to BaSO_4+Cu(NO_3)_2\\
Cu(NO_3)_2+2KOH\to Cu(OH)_2+2KNO_3\\
2Cu+O_2\xrightarrow{t^o}2CuO\\
CuO+H_2\xrightarrow{t^o}Cu+H_2O\\
Cu+2FeCl_3\to CuCl_2+2FeCl_2\\
FeCl_2+2KOH\to Fe(OH)_2+2KCl\\
4Fe(OH)_2+O_2+2H_2O\xrightarrow{t^o}4Fe(OH)_3\\
2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\
Fe_2O_3+2Al\xrightarrow{t^o}2Fe+Al_2O_3
\end{array}\)

\(\left(\sqrt{3}\right)^2=3< 49=7^2\\ \Rightarrow2+\sqrt{3}< 2+7=9\\ \Rightarrow\sqrt{2+\sqrt{3}}< \sqrt{9}=3\\ \Rightarrow1+\sqrt{2+\sqrt{3}}< 1+3=4\\ \Rightarrow\sqrt{1+\sqrt{2+\sqrt{3}}}< \sqrt{4}=2\)

`a)`

`n_{Al}={2,7}/{27}=0,1(mol)`

Bảo toàn e: `3n_{Al}=2n_{SO_2}`

`->n_{SO_2}=3/2n_{Al}=0,15(mol)`

`->V_{SO_2}=0,15.22,4=3,36(l)`

`b)`

`SO_2+H_2O`  $\leftrightharpoons$  `H_2SO_3`

Theo PT: `n_{H_2SO_3}=n_{SO_2}=0,15(mol)`

`->C_{M\ H_2SO_3}={0,15}/{0,5}=0,3M`

`(1)4Na+O_2`  $\xrightarrow{t^o}$  `2Na_2O`

`(2)Na_2O+H_2O->2NaOH`

`(3)CO_2+2NaOH->Na_2CO_3+H_2O`

`(4)Na_2CO_3+BaCl_2->2NaCl+BaCO_3`

`(5)NaCl+AgNO_3->AgCl+NaNO_3`

`(6)NaOH+HCl->NaCl+H_2O`

`(7)NaCl+H_2SO_4`  $\xrightarrow{250^oC}$  `NaHSO_4+HCl`

`(8)CuO+2HCl->CuCl_2+H_2O`

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

`(10)Cu(OH)_2` không ra được `Cu`

`(11)Cu+Cl_2`  $\xrightarrow{t^o}$  `CuCl_2`

`n_{Zn}={6,5}/{65}=0,1(mol)`

`n_{NaOH}=0,2.0,5=0,1(mol)`

`Zn+H_2SO_4->ZnSO_4+H_2`

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

Theo PT: `\sum n_{H_2SO_4}=0,1+1/2.0,1=0,15(mol)`

`->V_{dd\ H_2SO_4}={0,15}/1=0,15(l)`

\(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\left(x\ne-1;x\ne2\right)\\ \Leftrightarrow x-2-5\left(x+1\right)=-15\\ \Leftrightarrow x-2-5x-5=-15\\ \Leftrightarrow-4x=-8\\ \Leftrightarrow x=2\left(\text{loại}\right)\\ \text{Vậy }S=\varnothing\\ \dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\left(x\ne\pm2\right)\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)-x\left(x+2\right)=2-5x\\ \Leftrightarrow x^2-3x+2-x^2-2x=2-5x\\ \Leftrightarrow0x=0\\ \Leftrightarrow x\in R\\ \text{Vậy }S=R\backslash\left\{-2;2\right\}\\ \dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\left(x\ne0;x\ne\pm5\right)\\ \Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x+25\\ \Leftrightarrow2x^2+20x+50-x^2+10x-25=x+25\\ \Leftrightarrow x^2+29x=0\\ \Leftrightarrow x\left(x+29\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(\text{loại}\right)\\x=-29\end{matrix}\right.\\ \text{Vậy }S=\left\{-29\right\}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\\ =1+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...+\left(-\dfrac{1}{2021}+\dfrac{1}{2021}\right)-\dfrac{1}{2022}\\ =1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(ĐK:x\ne\pm3;x\ne0\\ M=\dfrac{\left(x+3\right)^2-\left(2x+5\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{1-x-3}{x\left(x+3\right)}\\ M=\dfrac{x^2+6x+9-2x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x+3\right)}{-\left(x+2\right)}\\ M=\dfrac{x^2+4x+4}{x-3}\cdot\dfrac{x}{-\left(x+2\right)}=\dfrac{x\left(x+2\right)^2}{3\left(3-x\right)}\\ M=\dfrac{x\left(x+2\right)^2}{\left(3-x\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3-x}\)