a)

Khí thoát ra: CH₄

\(\%V_{CH_4} = \dfrac{6,72}{16,8}.100\% = 40\%\\ \%V_{C_2H_4} = 100\% - 40\% = 60\%\)

b)

\(n_{C_2H_4} = \dfrac{16,8-6,72}{22,4} = 0,45(mol)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow C_{M_{Br_2}} = \dfrac{0,45}{2} = 0,225M\\ c) n_{C_2H_4Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow n_{C_2H_4Br_2} = 0,45.188 = 84,6(gam)\)

Bài 4:

a) n(hỗn hợp khí)= 16,8/22,4=0,75(mol)

- Khí thoát ra là khí CH4.

=> nCH4=6,72/22,4=0,3(mol)

nC2H4=0,75-0,3=0,45(mol)

- Số mol tỉ lệ thuận với thể tích.

%V(CH4)=%nCH4= (0,3/0,75).100=40%

=> %V(C2H4)=100% - 40%=60%

b) PTHH: C2H4 + Br2 -> C2H4Br2

nC2H4Br2= nBr2=nC2H4=0,45(mol)

=>VddBr2= 0,45/2=0,225(l)

c) mC2H4Br2=0,45. 188= 84,6(g)

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, \(n_{C_2H_4Br_2}=\dfrac{1,88}{188}=0,01\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,01\left(mol\right)\Rightarrow V_{C_2H_4}=0,01.22,4=0,224\left(l\right)\)

\(\Rightarrow V_{CH_4}=20-0,224=19,776\left(l\right)\)

c, \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,224}{20}.100\%=1,12\%\\\%V_{CH_4}=98,88\%\end{matrix}\right.\)

d, \(n_{Br_2}=n_{C_2H_4Br_2}=0,01\left(mol\right)\Rightarrow C_{M_{Br_2}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\)

a) \(n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2\left(mol\right)\)

PTHH: C₂H₄ + Br₂ ---> C₂H₄Br₂
             0,2<---0,2<------0,2

b) \(\left\{{}\begin{matrix}V_{C_2H_4}=0,2.24,79=4,958\left(l\right)\\V_{CH_4}=20-4,958=15,042\left(l\right)\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{4,958}{20}.100\%=24,79\%\\\%V_{CH_4}=100\%-24,79\%=75,21\%\end{matrix}\right.\)

d) \(V_{\text{dd}Br_2}=\dfrac{0,2}{0,2}=1\left(l\right)\)

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, \(n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2\left(mol\right)\)

\(n_{C_2H_4}=n_{C_2H_4Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

\(\Rightarrow V_{CH_4}=20-4,48=15,52\left(l\right)\)

c, \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{4,48}{20}.100\%=22,4\%\\\%V_{CH_4}=77,6\%\end{matrix}\right.\)

d, \(n_{Br_2}=n_{C_2H_4Br_2}=0,2\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,2}{0,2}=1\left(l\right)\)

a) \(V_{CH_4}=0,6\left(l\right)\)

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,6}{1,2}.100\%=50\%\\\%V_{C_2H_4}=100\%-50\%=50\%\end{matrix}\right.\)

b) \(n_{C_2H_4}=\dfrac{1,2-0,6}{24}=0,025\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂ 

         0,025-->0,025

=> \(m_{Br_2}=0,025.160=4\left(g\right)\)

c) 

\(n_{CH_4}=\dfrac{0,6}{24}=0,025\left(mol\right)\)

=> n_H = 0,025.4 = 0,1 (mol)

\(n_{Cl_2}=\dfrac{0,72}{24}=0,03\left(mol\right)\)

=> n_{Cl(thế H)} = 0,03 (mol)

Do n_H > n_{Cl(thế H)}

=> H không bị thế hoàn toàn bởi Cl

=> n_HCl = 0,03 (mol)

=> m_HCl = 0,03.36,5 = 1,095 (g)

nCO2=0,4(mol)

a) PTHH: 2 NaOH + CO2 -> Na2CO3 + H2O

0,8_________0,4________0,4(mol)

=> mNaOH=0,8.40=32(g) 

=>C%ddNaOH=(32/200).100=16%

b) mddNa2CO3=mddNaOH+mCO2=200+0,4.44=217,6(g)

mNa2CO3=106.0,4=42,4(g)

=>C%ddNa2CO3=(42,4/217,6).100=19,485%

Chúc em học tốt!

n_CO2=8,96/22,4=0,4mol

a/ CO₂+2NaOH→Na₂CO₃+H₂O

   0,4      0,8           0,4         0,4

m_NaOH=0,8.40=32g

C%dd_NaOH=m_ct/_mdd.100%=32/200.100%=16%

b/m_CO2=0,4.44=17,6g

Theo định luật bảo toàn khối lượng:

m_CO2+m_NaOH=m_Na2CO3

17,6g+200g=217,6g

m_Na2CO3=0,4.106=42,4g

C%ddNa2CO3=m_ct/mdd.100%=42,4/217,6.100=19,4852g

\(a)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{80.10\%}{160} = 0,05(mol)\\ \Rightarrow \%m_{C_2H_4} = \dfrac{0,05.28}{2}.100\% = 70\%\\ \%m_{CH_4} = 100\% - 70\% = 30\%\)

\(b)\) Sản phẩm : Đibrom etan

\(n_{C_2H_4Br_2} = n_{Br_2} = 0,05(mol)\\ \Rightarrow m_{C_2H_4Br_2} = 0,05.188 = 9,4\ gam\)

\(n_{hh}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(m_{Br_2}=200\cdot\dfrac{20}{100}=40\left(g\right)\)

\(n_{Br_2}=\dfrac{40}{160}=0.25\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(0.25........0.25..........0.25\)

\(\)\(n_{C_2H_4}=n_{hh}=0.25\left(mol\right)\)

=> Sai đề

\(n_{hh}=6,72:22,4=0,3mol\\ C_2H_2+2Br_2->C_2H_2Br_4\\ C_2H_4+Br_2->C_2H_2Br_2\\ n_{Br_2}=0,4mol\\ n_{C_2H_2}=a;n_{C_2H_4}=b\\ a+b=0,3\\ 2a+b=0,4\\ a=0,2;b=0,1\\ \%V_{C_2H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\ \%V_{C_2H_4}=33,33\%\)

1/2 hỗn hợp có 0,1 mol C2H2 và 0,05mol C2H4

\(BT.C:n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,3mol\\ n_{CaCO_3}=n_{CO_2}=0,3\\ m_{KT}=0,3.100=30g\)