a) \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b) \(CH_4+Cl_2\underrightarrow{\text{ánh sáng}}CH_3Cl+HCl\)

c) \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

d) \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

e) \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

f) \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Câu 1: C

Câu 2: D

Câu 3: D

Câu 4: B

Câu 5: D

Câu 6: C

Câu 7: C

Câu 8: A

Có kết tủa trắng xuất hiện: \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

Câu 9: A

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

            0,2---------------------->0,2

\(\Rightarrow m_{kt}=0,2.100=20\left(g\right)\)

Câu 10: Không có đáp án đúng

\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

0,5--------------->1

\(\Rightarrow V_{CO_2}=1.22,4=22,4\left(l\right)\)

\(M_X=17.2=34\left(g/mol\right)\)

Quy đổi hh \(X\left\{{}\begin{matrix}CH_4\\C_2H_4\\C_3H_4\\C_4H_4\end{matrix}\right.\rightarrow C_xH_4\left(1< x< 4\right)\)

\(m_X=34.0,075=2,55\left(g\right)\)

Mà \(n_H=4n_X=0,3\left(mol\right)\Rightarrow n_C=\dfrac{2,55-0,3}{12}=0,1875\left(mol\right)\)

BTNT: \(\left\{{}\begin{matrix}n_{CO_2}=n_C=0,1875\left(mol\right)\\n_{H_2O}=\dfrac{1}{2}n_H=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m=m_{CO_2}+m_{H_2O}=0,1875.44+0,15.18=10,95\left(g\right)\)

Chọn D

31) 

BTNT C: \(n_C=n_{CO_2}=n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\)

Ta có: \(m_{gi\text{ảm}}=m_{BaCO_3}-m_{CO_2}-m_{H_2O}\)

\(\Rightarrow59,1-0,3.44-m_{H_2O}=38,7\Leftrightarrow m_{H_2O}=7,2\left(g\right)\\ \Rightarrow n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,8\left(mol\right)\)

Ta có: \(n_{H_2O}>n_{CO_2}\left(0,4>0,3\right)\Rightarrow X\) thuộc dãy đồng đẳng ankan

Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

Ta có: \(n_X=n_{H_2O}-n_{CO_2}=0,1\left(mol\right)\)

Mà \(m_X=0,3.12+0,8=4,4\left(g\right)\)

\(\Rightarrow M_X=\dfrac{4,4}{0,1}=44\left(g/mol\right)\\ \Rightarrow14n+2=44\Leftrightarrow n=3\left(t/m\right)\)

Vậy X là C₃H₈ => Chọn C

a) Vì \(X+HCl\rightarrow H_2\uparrow\) `=> X` chứa kim loại dư

PTHH:

\(Mg+Cl_2\xrightarrow[]{t^o}MgCl_2\\ Zn+Cl_2\xrightarrow[]{t^o}ZnCl_2\\ 2Al+3Cl_2\xrightarrow[]{t^o}2AlCl_3\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\\ ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2\downarrow+2NaCl\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\\ Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\\ Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

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

BTNT Mg: \(n_{Mg}=n_{Mg\left(OH\right)_2}=\dfrac{5,8}{58}=0,1\left(mol\right)\)

\(\Rightarrow m_{Zn}+m_{Al}=14,85-0,1.24=12,45\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\Rightarrow65a+27b=12,45\left(1\right)\)

BTKL: \(m_{Cl_2}=29,05-14,85=14,2\left(g\right)\)

\(\Rightarrow n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)

Quá trình oxi hóa:

\(Mg^0\rightarrow Mg^{+2}+2e\)

0,1--------------->0,2

\(Zn^0\rightarrow Zn^{+2}+2e\)

a---------------->2a

\(Al^0\rightarrow Al^{+3}+3e\)

b-------------->3b

Quá trình khử:

\(Cl_2^0+2e\rightarrow2Cl^{-1}\)

0,2->0,4

\(2H^{+1}+2e\rightarrow H_2^0\)

            0,4<---0,2

`BTe: 0,2 + 2a + 3b = 0,4 + 0,4 => 2a + 3b = 0,6 (2)`

Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\left(TM\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{14,85}.100\%=16,16\%\\\%m_{Zn}=\dfrac{0,15.65}{14,85}.100\%=65,66\%\\\%m_{Al}=100\%-16,16\%-65,66\%=18,18\%\end{matrix}\right.\)

33)

a) \(C_2H_6+Cl_2\xrightarrow[]{a/s}C_2H_5Cl+HCl\)

b) \(CH_2=CH-CH_3\xrightarrow[]{t^o,xt}\left[{}\begin{matrix}CH\equiv C-CH_3+H_2\\CH_2=C=CH_2+H_2\end{matrix}\right.\)

c) \(CH_2=CH-CH_3+H_2\xrightarrow[]{t^o,xt}CH_3-CH_2-CH_3\)

d) \(CH_3-C\equiv CH+AgNO_3+NH_3\rightarrow CH_3-C\equiv CAg\downarrow+NH_4NO_3\)

e) \(CH_3-CH_2-CH_3+Cl_2\xrightarrow[]{a/s}CH_3-CH_2-CH_3Cl+HCl\)

f) \(CH_2=CH-CH_3+H_2O\xrightarrow[]{t^o,xt}\left[{}\begin{matrix}CH_3-CH\left(OH\right)-CH_3\left(sp.ch\text{ính}\right)\\OH-CH_2-CH_2-CH_3\left(sp.ph\text{ụ}\right)\end{matrix}\right.\)

g) \(CH\equiv CH+H_2O\xrightarrow[]{t^o,xt}CH_3CHO\)

h) \(CH_2=CH-CH_3+HCl\xrightarrow[]{t^o,xt}\left[{}\begin{matrix}CH_3-CHCl-CH_3\left(sp.ch\text{ính}\right)\\CH_2Cl-CH_2-CH_3\left(sp.ph\text{ụ}\right)\end{matrix}\right.\)

(1) \(CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4\uparrow+Na_2CO_3\)

(2) \(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\)

(3) \(C_2H_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}C_2H_4\)

(4) \(C_2H_4+HBr\xrightarrow[]{t^o,xt}C_2H_5Br\)

(5) \(2CH\equiv CH\xrightarrow[]{t^o,xt}CH_2=CH-C\equiv CH\)

(6) \(C_2H_2+2H_2\xrightarrow[]{Ni,t^o}C_2H_6\)

27) Chất làm mất màu là etilen, axetilen, but-1-en, pent-2-in

`=>` Có 4 chất

Chọn C

28) \(C_3H_8+5O_2\xrightarrow[]{t^o}3CO_2+4H_2O\)

       0,05--->0,25

`=> V_{O_2} = 0,25.22,4 = 5,6 (l)`

Chọn B

29) \(CH_2=CH_2+H_2O\xrightarrow[H_2SO_{4\left(\text{đ}\right)}]{t^o}CH_3-CH_2-OH\)

Chọn A

a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{200.8\%}{160}=0,1\left(mol\right);n_{hh}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

             0,1<-----0,1

b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,3}.100\%=33,33\%\\\%V_{CH_4}=100\%-33,33\%=66,67\%\end{matrix}\right.\)

c) \(n_{CH_4}=0,3-0,1=0,2\left(mol\right)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

             0,2--->0,4

            \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

              0,1--->0,3

\(\Rightarrow V_{O_2}=\left(0,3+0,4\right).24,79=17,353\left(l\right)\)

22) \(n_{C_3H_6}=\dfrac{2,52}{42}=0,06\left(mol\right)\)

PTHH: \(2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\)

             0,06------------->0,18---->0,18

\(\Rightarrow m_{t\text{ăng}}=0,18.\left(44+18\right)=11,16\left(g\right)\)

Chọn B

23) Propan là \(CH_3-CH_2-CH_3\). Chọn D

24) Chất không tạo kết tủa là \(CH_3-C\equiv C-CH_3\) do không phải là ank-1-in

\(CH\equiv CH+2AgNO_3+2NH_3\rightarrow CAg\equiv CAg\downarrow+2NH_4NO_3\\ CH_3-C\equiv CH+AgNO_3+NH_3\rightarrow CH_3-C\equiv CAg\downarrow+NH_4NO_3\\ CH\equiv C-CH=CH_2+AgNO_3+NH_3\rightarrow CAg\equiv C-CH=CH_2\downarrow+NH_4NO_3\)

Chọn A

25) BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,9\left(mol\right)\\n_H=2n_{H_2O}=1,2\left(mol\right)\end{matrix}\right.\Rightarrow m=0,9.12+1,2=12\left(g\right)\)

Chọn C

26) \(CH_4+4Cl_2\xrightarrow[]{a/s}CCl_4+4HCl\)

Chọn D