$n_{NaOH}=0,5.0,06=0,03(mol)$

$\sum n_{H_2SO_4}=0,3.0,25=0,075(mol)$

$R+H_2SO_4\to RSO_4+H_2(1)$

$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O(2)$

Theo PT: $n_{H_2SO_4(2)}=\frac{1}{2}n_{NaOH}=0,015(mol)$

$\to n_{H_2SO_4(1)}=0,075-0,015=0,06(mol)$

Theo PT: $n_R=n_{H_2SO_4(1)}=0,06(mol)$

$\to M_R=\frac{1,44}{0,06}=24(g/mol)$

Vậy $R$ là magie 

$V_{C_2H_5OH\,nguyên\,chất}=\frac{5.40}{100}=2(l)=2000(ml)$

$\to m_{C_2H_5OH}=2000.0,8=1600(g)$

Vì $H=92\%$

$\to n_{C_2H_5OH(pứ)}=\frac{1600.92\%}{46}=32(mol)$

$C_2H_5OH+O_2\xrightarrow{\rm men\,giấm}CH_3COOH+H_2O$

Theo PT: $n_{CH_3COOH}=n_{C_2H_5OH}=32(mol)$

$\to m_{CH_3COOH}=32.60=1920(g)$

$a)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$b)4H_2+Fe_3O_4\xrightarrow{t^o}3Fe+4H_2O$

$c)4P+5O_2\xrightarrow{t^o}2P_2O_5$

$d)2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2$

$e)S+O_2\xrightarrow{t^o}SO_2$

$f)PbO+H_2\xrightarrow{t^o}Pb+H_2O$

$a\big)$

$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$

$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$

Theo PT: $n_{Zn}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$

$\to \%m_{Zn}=\frac{0,2.65}{21,1}.100\%\approx 61,61\%$

$\to \%m_{ZnO}=100-61,61=38,39\%$

$b\big)$

$n_{ZnO}=\frac{21,1-0,2.65}{81}=0,1(mol)$

Theo PT: $\sum n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,6(mol)$

$\to C_{M_{CH_3COOH}}=\dfrac{0,6}{\frac{200}{1000}}=3M$

$a\big)$

Bảo toàn C: $n_C=n_{CO_2}=\frac{2,24}{22,4}=0,1(mol)$

Bảo toàn H: $n_H=2n_{H_2O}=2.\frac{2,7}{18}=0,3(mol)$

$\to n_O=\frac{2,3-0,1.12-0,3}{16}=0,05(mol)$

$\to n_C:n_H:n_O=0,1:0,3:0,05=2:6:1$

$\to$ CT nguyên là $(C_2H_6O)_n$

Mà $M_A=46(g/mol)$

$\to (12.2+6+16).n=46$

$\to n=1$

Vậy CTPT của A là $C_2H_6O$

$b\big)$

$CH_3-CH_2-OH$

$CH_3-O-CH_3$

$n_{Mg}+n_{Fe}=n_X=0,3$

Mà $n_{Mg}:n_{Fe}=2:1$

$\to n_{Mg}=2n_{Fe}$

$\to 2n_{Fe}+n_{Fe}=0,3$

$\to n_{Fe}=0,1(mol)$

$\to n_{Mg}=0,2(mol)$

$\to m_{hỗn\,hợp}=0,2.24+0,1.56=10,4(g)$

\(P=\dfrac{\left[\left(x-y\right)\left(1+xy\right)\right]\left[\left(x+y\right)\left(1-xy\right)\right]}{\left(1+x^2\right)^2\left(1+y^2\right)^2}\)

Áp dụng BĐT Cosi ta có:

\(\left(x-y\right)\left(1+xy\right)\le\dfrac{\left(x-y\right)^2+\left(1+xy\right)^2}{2}=\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{2}\\ \left(x+y\right)\left(1-xy\right)\le\dfrac{\left(x+y\right)^2+\left(1-xy\right)^2}{2}=\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{2}\)

\(\to P\le\dfrac{\left(1+x^2\right)^2\left(1+y^2\right)^2}{4\left(1+x^2\right)^2\left(1+y^2\right)^2}=\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

$a\big)$

Bảo toàn C: $n_C=n_{CO_2}=\frac{6,72}{22,4}=0,3(mol)$

Bảo toàn H: $n_H=2n_{H_2O}=2.\frac{5,4}{18}=0,6(mol)$

$\to n_C:n_H=0,3:0,6=1:2$

$\to$ Công thức nguyên là $(CH_2)_n$

Mà $M_A=21.2=42(g/mol)$

$\to (12+2).n=42$

$\to n=3$

Vậy CTPT của A là $C_3H_6$

$b\big)CH_2=CH-CH_3$

21.

$2KClO_3\xrightarrow{t^o,MnO_2}2KCl+3O_2$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$\to D$

22.

$2KClO_3\xrightarrow{t^o}2KCl+3O_2$

$\to A$

23. $D$

24. $C$

25.

$n_{KClO_3}=\frac{12,25}{122,5}=0,1(mol)$

$2KClO_3\xrightarrow{t^o,MnO_2}2KCl+3O_2$

Theo PT: $n_{O_2}=\frac{3}{2}n_{KClO_3}=0,15(mol)$

$\to V_{O_2}=0,15.22,4=3,36(l)$

$\to B$

$C_2H_4+Br_2\to C_2H_4Br_2$

$C_2H_2+2Br_2\to C_2H_2Br_4$

Đặt $n_{C_2H_4}=x(mol);n_{C_2H_2}=y(mol)$

$\to \begin{cases} x+y=\frac{2,8}{22,4}=0,125\\ x+2y=n_{Br_2}=\frac{24}{160}=0,15 \end{cases}$

$\to \begin{cases} x=0,1\\ y=0,025 \end{cases}$

$\to\begin{cases} \%V_{C_2H_4}=\frac{0,1}{0,125}.100\%=80\%\\ \%V_{C_2H_2}=100-80=20\% \end{cases}$