\(\left(1\right)\left(C_6H_{10}O_5\right)_n+nH_2O\underrightarrow{H^+t^o}C_6H_{12}O_6\)

\(\left(2\right)C_6H_{12}O_6\underrightarrow{lênmen}2C_2H_5OH+2CO_2\uparrow\)

\(\left(3\right)C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)

\(\left(4\right)CH_3COOH+C_2H_5OH\underrightarrow{t^o,xt}CH_3COOC_2H_5+H_2O\)

\(\left(5\right)CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\)

\(\left(6\right)HOCH_2\left[CHOH\right]_4CHO+Br_2+H_2O\rightarrow HOCH_2\left[CHOH\right]_4COOH+2HBr\)

\(\left(7\right)C_2H_5OH\underrightarrow{H_2SO_4,170^oC}C_2H_4+H_2O\)

\(\left(8\right)2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\)

\(\left(9\right)CH_3COOC_2H_5+NaOH\rightarrow C_2H_5OH+CH_3COONa\)

\(n_{hh}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\)

Gọi số mol của \(C_2H_4\) là: \(a\)

Gọi số mol của \(C_2H_2Br_4\) là: \(b\)

\(PTHH:\\ +)C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\+)C_2H_4+Br_2\rightarrow C_2H_4Br_2 \)

\(\left\{{}\begin{matrix}a+b=0,3\\a+2b=0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\%V_{C_2H_4}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_2}=100\%-66,67\%=33,33\%\)

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

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

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

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

$a\big)$

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

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

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

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

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

 0,2                                                                0,2     ( mol )

\(m_{Zn}=0,2.65=13g\)

\(\%m_{Zn}=\dfrac{13}{21,1}.100=61,61\%\)

\(\%m_{ZnO}=100\%-61,61\%=38,39\%\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,2               0,4                                                  ( mol )

\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1mol\)

\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)

 0,1              0,2                                                        ( mol )

\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)

Hợp chất A gồm C,H và có thể có O

\(n_C=n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(n_H=2.n_{H_2O}=2.\dfrac{2,7}{18}=0,3mol\)

\(n_O=\dfrac{2,3-\left(0,1.12+0,3.1\right)}{16}=0,05mol\)

\(\Rightarrow CT:C_xH_yO_z\)

\(x:y:z=0,1:0,3:0,05=2:6:1\)

\(\Rightarrow CPPT:C_2H_6O\)

\(\left(C_2H_6O\right)n=46\)

\(\Rightarrow n=1\)

\(CTCT:CH_3-CH_2-OH\)

$a\big)$

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 $C_2H_6O$

$b\big)$

$CH_3-CH_2-OH$

$CH_3-O-CH_3$

$a\big)$

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 $C_3H_6$

$b\big)CH_2=CH-CH_3$

$C_2H_4+Br_2\to C_2H_4Br_2$

$C_2H_2+2Br_2\to C_2H_2Br_4$

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

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

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

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

CH4+Cl2-as->CH3Cl+HCl

CaC2+2H2O->Ca(OH)2+C2H2

C2H2+5\2O2-to>2CO2+H2O

C2H2+2H2-xt->C2H6

bài 5

C2H5OH+3O2-to>2CO2+3H2O

0,5------------1,5--------1 mol

n C2H5OH=0,5 mol

=>VCo2=1,5.22,4=33,6l

b)Vkk=1,5.22,4.5=168l

Thi hẽm? :))

Do đốt cháy A sinh ra sản phẩm chứa các nguyên tố C, H, O; A gồm 2 nguyên tố

=> A gồm C, H

\(\left\{{}\begin{matrix}n_{CO_2}=2\left(mol\right)\\n_{H_2O}=2\left(mol\right)\end{matrix}\right.\)

Bảo toàn O: \(2.n_{O_2}=2.n_{CO_2}+n_{H_2O}\)

=> n_O2 = 3 (mol)

=> V_O2 = 3.22,4 = 67,2 (l)