C₂H₄+Br₂->C₂H₄Br₂

0,05----0,05

n _Br2=\(\dfrac{8}{160}\)=0,05 mol

=>%V_C2H4=\(\dfrac{0,05.22,4}{5,6}.100=20\%\)

=>%V_CH4=80%

c)CH₄+2O₂-to>CO₂+2H₂O

1.10^-3----2.10^-3 mol

C₂H₄+3O₂-to>2CO₂+2H₂O

2,5.10^-4-7,5.10^-4 mol

n hh=\(\dfrac{0,028}{22,4}\)=1,25.10^-3 mol

=>n _C2H4=2,5.10^-4 mol

=>n _CH4=1.10^-3 mol

=>V_O2=(2.10^-3+7,5.10^-4).22,4=0,0616l

\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)

\(\Rightarrow n_{etilen}=n_{Br_2}=0,05mol\)

\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)

\(\Rightarrow n_{metan}=n_{hh}-n_{etilen}=0,25-0,05=0,2mol\)

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

b)\(\%V_{metan}=\dfrac{0,2}{0,25}\cdot100\%=80\%\)

\(\%V_{etilen}=100\%-80\%=20\%\)

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

b) Ta có: \(n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,07\cdot22,4}{1,72}\cdot100\%\approx91,16\%\) 

\(\Rightarrow\%V_{CH_4}=8,84\%\)

a) \(C_2H_4 + Br_2 \to C_2H_4Br_2\)

b)

\(n_{C_2H_4} = n_{Br_2} = \dfrac{5,6}{160} = 0,035(mol)\\ \%V_{C_2H_4} = \dfrac{0,035.22,4}{0,86}.100\% = 91,16\%\\ \%V_{CH_4} = 100\% - 91,16\% = 8,84\%\)

a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{6,4}{160}=0,04\left(mol\right);n_{hh}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

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

             0,04<--0,04

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04}{0,6}.100\%=6,67\%\\\%V_{CH_4}=100\%-6,67\%=93,33\%\end{matrix}\right.\)

b) \(n_{CH_4}=0,6-0,04=0,56\left(mol\right)\)

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

              0,56----------->0,56

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

              0,04----------->0,08

\(\Rightarrow V_{CO_2}=\left(0,08+0,56\right).22,4=14,336\left(l\right)\)

a, nBr2 = 8/160 = 0,05 (mol)

PTHH: C2H4 + Br2 -> C2H4Br2

Mol: 0,05 <--- 0,05 <--- 0,05

Vhh khí = 2,8/22,4 = 0,125 (mol)

%VC2H4 = 0,05/0,125 = 40%

%CH4 = 100% - 40% = 60%

b, nCH4 = 0,125 - 0,05 = 0,075 (mol)

PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

Mol: 0,05 ---> 0,15

CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,075 ---> 0,15

Vkk = (0,15 + 0,15) . 5 . 22,4 = 33,6 (l)

ta có :

nBr2=\(\dfrac{16}{160}=0,1mol\)

C2H4+Br2->C2H4Br2

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

=>VC2H4=0,1.22,4=2,24l

=>VCH4=3,36l->n CH4=0,15 mol

->%VC2H4=\(\dfrac{2,24}{5,6}.100\)=40%

=>%VCH4=60%

c)

CH4+2O2-to>CO2+2H2O

0,15---------------0,15

C2H4+3O2--to>2CO2+2H2O

0,1--------------------0,2
=>m CaCO3=0,35.100=35g

a, PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

b, \(n_{Br_2}=\dfrac{5,6}{160}=0,035\left(mol\right)\)

Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,0175\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_2}=\dfrac{0,0175.22,4}{0,86}.100\%\approx45,58\%\)

\(\Rightarrow\%V_{CH_4}\approx54,42\%\)

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Giả sử: \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x+y=\dfrac{1,344}{22,4}=0,06\left(1\right)\)

Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y\left(mol\right)\)

⇒ x + 2y = 0,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,02}{0,06}.100\%\approx33,33\%\\\%\text{ }V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Ta có: 1/2 hỗn hợp khí gồm: 0,01 mol C₂H₄ và 0,02 mol C₂H₂.

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

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

Theo PT: \(n_{CO_2}=2n_{C_2H_4}+2n_{C_2H_2}=0,06\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{cr}=m_{CaCO_3}=0,06.100=6\left(g\right)\)

Bạn tham khảo nhé!

a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

\(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,025\left(mol\right)\Rightarrow m_{C_2H_2}=0,025.26=0,65\left(g\right)\)

b, \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,025.22,4}{33,6}.100\%\approx1,67\%\\\%V_{CH_4}\approx98,33\%\end{matrix}\right.\)