a) $2Mg+O^2\stackrel{t^o}{\to }MgO$ - pư hóa hợp

b) $F{e}^3{O}^4+4H^2\stackrel{t^o}{\to }3Fe+4H^{2}O$ - pư thế

c) $Fe+H^{2}S{O}^{4\left(l\right)}\to FeS{O}^4+H^2$ - pư thế

d) $2KCl{O}^3\stackrel{t^o}{\to }2KCl+3O^2$ - pư phân hủy

PT: \(2C_3H_8O_3+7O_2\underrightarrow{t^o}6CO_2+8H_2O\) (1) 

\(C_2H_4O_2+2O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(C_3H_6O_3+3O_2\underrightarrow{t^o}3CO_2+3H_2O\)

\(C_6H_{12}O_6+6O_2\underrightarrow{t^o}6CO_2+6H_2O\)

Coi hh gồm: C₃H₈O₃ và CH₂O (vì C₂H₄O₂, C₃H₆O₃ và C₆H₁₂O₆ đều có CTĐGN là CH₂O)

\(CH_2O+O_2\underrightarrow{t^o}CO_2+H_2O\)

Ta có: \(n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\)

\(n_{H_2O}=\dfrac{19,8}{18}=1,1\left(mol\right)\)

\(\Rightarrow n_{H_2O}-n_{CO_2}=n_{C_3H_8O_3}=0,1\left(mol\right)\)

BTNT C, có: \(3n_{C_3H_8O_3}+n_{CH_2O}=n_{CO_2}\Rightarrow n_{CH_2O}=0,7\left(mol\right)\)

⇒ m = m_C3H8O3 + m_CH2O = 0,1.92 + 0,7.30 = 30,2 (g)

Theo ĐLBT KL, có: m_A + m_O2 = m_CO2 + m_H2O

⇒ m_O2 = 44 + 19,8 - 30,2 = 33,6 (g) \(\Rightarrow n_{O_2}=\dfrac{33,6}{32}=1,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,05.22,4=23,52\left(l\right)\)

I)

1) 

\(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ CH_2=CH-CH=CH_2+2H_2\xrightarrow[]{Ni,t^o}CH_3-CH_2-CH_2-CH_3\\ CH_3-CH_2-CH_2-CH_3\xrightarrow[]{cracking}CH_2=CH-CH_3+CH_4\)

2)

\(CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4+Na_2CO_3\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH_2\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

3)

\(2CH_4\xrightarrow[]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ nCH_2=CH-CH=CH_2\xrightarrow[]{t^o,p,xt}\left(-CH_2-CH=CH-CH_2-\right)_n\)

4) 

\(C_4H_8+H_2\xrightarrow[]{Ni,t^o}C_4H_{10}\\ C_4H_{10}\xrightarrow[]{cracking}CH_4+C_3H_6\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\)

II)

1)

\(CH\equiv C-CH_2-CH_3+2Br_2\rightarrow CHBr_2-CBr_2-CH_2-CH_3\\ CH_3-C\equiv C-CH_3+2Br_2\rightarrow CH_3-CBr_2-CBr_2-CH_3\\ CH\equiv C-CH_2-CH_3+AgNO_3+NH_3\rightarrow CAg\equiv C-CH_2-CH_3\downarrow+NH_4NO_3\)

2)

\(CH\equiv CH+2AgNO_3+2NH_3\rightarrow CAg\equiv CAg\downarrow+2NH_4NO_3\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

III)

1) \(n_C=n_{CO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(\Rightarrow n_H=\dfrac{0,88-0,06.12}{1}=0,16\left(mol\right)\Rightarrow n_{H_2O}=\dfrac{1}{2}n_{H_2O}=0,08\left(mol\right)\\ \Rightarrow n_{ankan}=n_{H_2O}-n_{CO_2}=0,02\left(mol\right)\\ \Rightarrow M_{ankan}=\dfrac{0,88}{0,02}=44\left(g/mol\right)\)

Đặt CTHH của ankan là C_nH_2n+2

=> 14n + 2 = 44 => n = 3

Vậy X là C₃H₈ \(\left(CTCT:CH_3-CH_2-CH_3:propan\right)\)

2) \(n_{H_2O}=\dfrac{1,62}{18}=0,09\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,18\left(mol\right)\)

\(\Rightarrow n_{CO_2}=n_C=\dfrac{1,62-0,18}{12}=0,12\left(mol\right)\\ \Rightarrow n_{ankin}=n_{CO_2}-n_{H_2O}=0,03\left(mol\right)\\ \Rightarrow nM_{ankin}=\dfrac{1,62}{0,03}=54\left(g/mol\right)\)

Đặt CTHH của ankin là C_nH_2n-2

=> 14n - 2 = 54 => n = 4

Vậy X là C₄H₆

CTCT: 

\(CH\equiv C-CH_2-CH_3:\) but-1-in

\(CH_3-C\equiv C-CH_3:\) but-2-in

3)

Sửa đề: 1,17 -> 11,7

\(\left\{{}\begin{matrix}n_{H_2O}=\dfrac{11,7}{18}=0,65\left(mol\right)\\n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\end{matrix}\right.\Rightarrow n_{H_2O}>n_{CO_2}\Rightarrow\) hh thuốc dãy đồng đẳng ankan

Ta có: \(n_{hh}=n_{H_2O}-n_{CO_2}=0,25\left(mol\right)\)

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,4\left(mol\right)\\n_H=2n_{H_2O}=1,3\left(mol\right)\end{matrix}\right.\Rightarrow m_{hh}=0,4.12+1,3=6,1\left(g\right)\)

\(\Rightarrow M_{hh}=\dfrac{6,1}{0,25}=24,4\left(g/mol\right)\)

Đặt CT chung của hh là C_nH_2n+2

=> 14n + 2 = 24,4 => n = 1,6 

=> Hai hiđrocacbon là CH₄ và C₂H₆

Ta có: \(\dfrac{m_M}{m_N}=\dfrac{2}{5}\Rightarrow5m_M-2m_N=0\left(1\right)\)

Mà: m_M + m_N = 1 (tấn) = 1000 (kg) (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_M=\dfrac{2000}{7}\left(kg\right)\\m_N=\dfrac{5000}{7}\left(kg\right)\end{matrix}\right.\)

\(\Rightarrow m_{Fe_2O_3}=m_M.60\%=\dfrac{1200}{7}\left(kg\right)\Rightarrow n_{Fe_2O_3}=\dfrac{\dfrac{1200}{7}}{160}=\dfrac{15}{14}\left(kmol\right)\)

\(\Rightarrow n_{Fe}=2n_{Fe_2O_3}=\dfrac{15}{7}\left(kmol\right)\)

\(m_{Fe_3O_4}=m_N.69,6\%=\dfrac{3480}{7}\left(kg\right)\Rightarrow n_{Fe_3O_4}=\dfrac{\dfrac{3480}{7}}{232}=\dfrac{15}{7}\left(kmol\right)\)

\(\Rightarrow n_{Fe}=3n_{Fe_3O_4}=\dfrac{45}{7}\left(kmol\right)\)

\(\Rightarrow m_{Fe}=\left(\dfrac{15}{7}+\dfrac{45}{7}\right).56=480\left(kg\right)\)

Bạn xem lời giải ở đây nhé.

https://olm.vn/hoi-dap/detail/7712766586263.html

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Có: O₂ hao hụt 40% → H% = 100 - 40 = 60%

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,4}{60\%}=\dfrac{2}{3}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=\dfrac{2}{3}.158\approx105,3\left(g\right)\)

Xăng, dầu đk nhỉ

\(M_A=5.2=10\left(g/mol\right)\)

Do các khí đo ở cùng điều kiện nhiệt độ và áp suất nên tỉ lệ thể tích cũng là tỉ lệ mol

Chọn \(\left\{{}\begin{matrix}n_{H_2}=17,5\left(mol\right)\\n_{N_2}=5\left(mol\right)\end{matrix}\right.\)

Gọi \(n_{N_2\left(p\text{ư}\right)}=a\left(mol\right)\left(0< a< 5\right)\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

             a---->3a---------->2a

Xét tỉ lệ: \(5< \dfrac{17,5}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂

Ta có: \(n_A=5+17,5+2a-a-3a=22,5-2a\left(mol\right)\)

Theo ĐLBTKL: \(m_A=5.28+17,5.2=175\left(g\right)\)

\(\Rightarrow M_A=\dfrac{175}{22,5-2a}=10\Leftrightarrow a=2,5\left(TM\right)\)

\(\Rightarrow H=\dfrac{2,5}{5}.100\%=50\%\)