2: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

4: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

CH4 + Cl2 -> (askt) CH3Cl + HCl 

Zn + 2 HCl -> ZnCl2 + H2 

CaCO3 -to-> CaO + CO2 

Nếu câu 3 là: CH2=CH2 + Br2 -> CHBr-BrCH

Tham khảo:

+) Dẫn lần lượt từng khí qua dd Ca(OH)2 dư, khí nào tạo kết tủa trắng là CO2:
PTHH: 

+) 2 khí còn lại cho qua dd Br2 dư, khí nào làm dd Br2 mất màu là C2H4:
PTHH: 
⇒ khí còn lại là CH4

Mình cảm ơn bạn nhaaaa

Gọi \(\left\{{}\begin{matrix}n_{CO}=a\left(mol\right)\\n_{C_nH_{2n+2}}=b\left(mol\right)\end{matrix}\right.\)

=> \(a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Bảo toàn C: \(a+bn=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Bảo toàn H: \(2bn+2b=\dfrac{2,7}{18}.2=0,3\left(mol\right)\)

=> a = 0,15; b = 0,05; n = 2

=> CTPT: C₂H₆

\(\left\{{}\begin{matrix}\%V_{C_2H_6}=\dfrac{0,05}{0,2}.100\%=25\%\\\%V_{CO}=\dfrac{0,15}{0,2}.100\%=75\%\end{matrix}\right.\)

\(n_{O_2}=\dfrac{20-14,24}{32}=0,18\left(mol\right)\\ Đặt:M\left(hoá.trị.x\right)\\ 4M+xO_2\rightarrow\left(t^o\right)2M_2O_x\\ n_{M_2O_x}=\dfrac{0,18.2}{x}=\dfrac{0,36}{x}\left(mol\right)\\ M_2O_x+2xHCl\rightarrow2MCl_x+xH_2O\\ n_{HCl}=\dfrac{0,36}{x}.2x=0,72\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,72}{0,8}=0,9\left(lít\right)\)

\(n_{H_2SO_4}=\dfrac{19,6\%.500.1,12}{98}=1,12\left(mol\right)\\n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ MgO+H_2SO_4 \rightarrow MgSO_4+H_2O\\ CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\\ TH1:axit.hết\\ n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\\ \Rightarrow n_{MgO}=\dfrac{18-0,1.10}{40}=0,2\left(mol\right)\\ n_{H_2SO_4\left(p.ứ\right)}=0,2+0,1=0,3\left(mol\right)< 1,12\left(mol\right)\\ \Rightarrow LoạiTH1\\ TH2:axit.dư\\ \Rightarrow\left\{{}\begin{matrix}40a+100b=18\\b=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{MgO}=\dfrac{0,2.40}{18}.100\approx44,444\%\Rightarrow\%m_{CaCO_3}\approx55,556\%\)

\(b,m_{ddB}=m_A+m_{ddH_2SO_4}-m_{CO_2}=18+500.1,12-0,1.44=573,6\left(g\right)\\ n_{H_2SO_4\left(dư\right)}=1,12-0,3=0,82\left(mol\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,82.98}{573,6}.100\approx14,01\%\\ C\%_{ddCaCl_2}=\dfrac{0,1.111}{573,6}.100\approx1,935\%\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{573,6}.100\approx3,312\%\)

\(n_{HCl}=0,4.2=0,8\left(mol\right)\\ Đặt:n_{CuO}=a\left(mol\right);n_{Fe_2O_3}=b\left(mol\right)\left(a,b>0\right)\\CuO+2HCl\rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O \\ \Rightarrow\left\{{}\begin{matrix}80a+160b=24\\2a+6b=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ C1:m_{muối}=135a+2.162,5b=135.0,1+325.0,1=46\left(g\right)\\ C2:n_{H_2O}=\dfrac{n_{HCl}}{2}=0,4\left(mol\right)=n_{O\left(trongH_2O\right)}=n_{O\left(trong.oxit\right)}\\ \Rightarrow m_{muối}=m_{CuO,Fe_2O_3}-m_{O\left(trong.oxit\right)}+m_{Cl^-}=24-0,4.16+0,8.35,5=46\left(g\right)\\ b,m_{CuO}=80a=8\left(g\right);m_{Fe_2O_3}=160b=16\left(g\right)\)

Làm vài bài rồi ra sân bay check in đây

CuO+2HCl->CuCl₂+H₂O

x--------2x

Fe₂O₃+6HCl->2FeCl₃+3H₂O

y--------------6y

Ta  có :

\(\left\{{}\begin{matrix}56x+160y=24\\2x+6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

=>m _CuO=0,1.56=5,6g

=>m _Fe2O3=0,1.160=16g

-> m _muối =0,1.135+0,2.162,5=46g

C2:n _H2O=0,4 mol

=>m _muối =24-0,4.16+0,8.35,5=46g

\(n_{CO_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right);n_{CH_4}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Đặt:n_{CH_4}=a\left(mol\right);n_{C_2H_4}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+2b=0,4\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow\%V_{CH_4}=\%n_{CH_4}=\dfrac{a}{0,3}.100\%=\dfrac{0,2}{0,3}.100\approx66,667\%\\ \Rightarrow\%V_{C_2H_4}\approx33,333\%\\ c,C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\\ \Rightarrow C_{MddBr_2}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

Chỗ kia chắc 200ml dung dịch Br₂ chứ 200gam thì cần cho thêm KLR á em

C₂H₄+Br₂-to>C₂H₄Br₂

x--------x mol

C₂H₂+2Br₂-to>C₂H₂Br₄

y----------2y mol

ta có :

\(\left\{{}\begin{matrix}x+y=0,3\\160x+320y=80\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=>%n_C2H2=\(\dfrac{0,1}{0,3}.100=33,3\%\)

=>%n_C2H4=100-33,3=66,7%

Thanks nhiều ạ

Tks :)