\(etan\left(C_2H_6\right):\)không phản ứng với \(H_2,Br_2,AgNO_3\text{/}NH_3,HCl,H_2O\)

\(eten\left(C_2H_4\right):\)phản ứng với \(H_2,Br_2,HCl,H_2O\)

\(C_2H_4+H_2\xrightarrow[t^o]{Ni}C_2H_6\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ C_2H_4+HCl\rightarrow C_2H_5Cl\\ C_2H_4+H_2O\xrightarrow[H^+,\text{men rượu}]{t^o}C_2H_5OH\)

\(etin\left(C_2H_2\right):\)phản ứng với \(H_2,Br_2,AgNO_3\text{/}NH_3,HCl,H_2O\)

\(\left(1\right)C_2H_2+H_2\xrightarrow[t^o]{Pd\text{/}PdCO_3}C_2H_4\\ \left(2\right)C_2H_4+2H_2\xrightarrow[t^o]{Ni}C_2H_6\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ C_2H_2+2AgNO_3+2NH_3\rightarrow C_2Ag_2\downarrow+2NH_4NO_3\)

\(C_2H_2+HCl\xrightarrow[HgCl_2]{t^o}C_2H_3Cl\)

\(C_2H_2+H_2O\xrightarrow[Hg^{2+}\text{/}H_2SO_4]{t^o}CH_3CHO\)

Gọi \(\left\{{}\begin{matrix}n_{C_6H_5OH}=a\left(mol\right)\\n_{CH_3OH}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH:

\(C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\)

a-------------------------------------->0,5a

\(CH_3OH+Na\rightarrow CH_3ONa+\dfrac{1}{2}H_2\)

b-------------------------------------->0,5b

Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}94a+32b=22\\0,5a+0,5b=0,15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\left(TM\right)\) hay \(\left\{{}\begin{matrix}n_{C_6H_5OH}=0,2\left(mol\right)\\n_{CH_3OH}=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{C_6H_5OH}=\dfrac{0,2.94}{22}.100\%=85,45\%\\\%m_{CH_3OH}=100\%-85,45\%=14,55\%\end{matrix}\right.\)

b) Trong 11 gam hỗn hợp có:

\(\left\{{}\begin{matrix}n_{C_6H_5OH}=\dfrac{11}{22}.0,2=0,1\left(mol\right)\\n_{CH_3OH}=\dfrac{11}{22}.0,1=0,05\left(mol\right)\end{matrix}\right.\)

PTHH: \(C_6H_5OH+3HNO_3\xrightarrow[H_2SO_4\left(đặc\right)]{}3H_2O+C_6H_2OH\left(NO_2\right)_3\)

               0,1--------------------------------------------->0,1

\(\rightarrow m_{C_6H_2OH\left(NO_2\right)_3}=0,1.229=22,9\left(g\right)\)

tham khảo:| Cộng đồng Học sinh Việt Nam - HOCMAI Forum

\(A.Na\)

PTHH: 

\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\uparrow\)

\(2C_3H_5\left(OH\right)_3+6Na\rightarrow2C_3H_5\left(ONa\right)_3+3H_2\uparrow\)

\(\begin{array} {l} 13)\\ n_{HCHO}=\dfrac{1,2}{30}=0,04(mol)\\ HCHO\xrightarrow{+AgNO_3/NH_3,t^o}4Ag\\ n_{Ag}=4n_{HCHO}=0,16(mol)\\ m=0,16.108=17,28(g)\\ \to A\\ 14)\\ X:C_nH_{2n}\\ n_{Br_2}=\dfrac{8}{160}=0,05(mol)\\ C_nH_{2n}+Br_2\to C_nH_{2n}Br_2\\ n_{C_nH_{2n}}=n_{Br_2}=0,05(mol)\\ M_{C_nH_{2n}}=14n=\dfrac{1,4}{0,05}=28(g/mol)\\ n=2\\ X:C_2H_4\\ \to A \end{array}\)

Đặt \(n_{C_3H_8O_3}=a\left(mol\right)\) \(\Rightarrow n_{C_2H_6O}=2a\left(mol\right)\)

Theo đề bài: \(92a+46\cdot2a=27,6\) \(\Leftrightarrow a=0,1\)

PTHH: \(C_2H_6O+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)

            \(C_3H_5\left(OH\right)_3+3Na\rightarrow C_3H_5\left(ONa\right)_3+\dfrac{3}{2}H_2\uparrow\)

Ta có: \(n_{H_2\left(lí.thuyết\right)}=\dfrac{1}{2}n_{etanol}+\dfrac{3}{2}n_{glixerol}=0,25\left(mol\right)\) \(\Rightarrow V_{H_2\left(thực\right)}=0,25\cdot22,4\cdot95\%=5,32\left(l\right)\)