m_tăng = m_C2H4

=> \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)

=> \(n_{CH_4}=\dfrac{8,96}{22,4}-0,2=0,2\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂  + 2H₂O

            0,2--->0,4

            C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

             0,2---->0,6

=> V_O2 = (0,4 + 0,6).22,4 = 22,4 (l)

=> V_kk = 22,4.5 = 112 (l)

Ta có: m _{dd Br2 tăng} = m_C2H4 = 2,8 (g)

\(\Rightarrow n_{C_2H_4}=\dfrac{2,8}{28}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{3,36}.100\%\approx66,67\%\\\%V_{CH_4}\approx33,33\%\end{matrix}\right.\)

Có: \(n_{CH_4}=\dfrac{3,36}{22,4}-0,1=0,05\left(mol\right)\)

⇒ m _hh = m_CH4 + m_C2H4 = 0,05.16 + 0,1.28 = 3,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{3,6}.100\%\approx22,22\%\\\%m_{C_2H_4}\approx77,78\%\end{matrix}\right.\)

1. \(n_{Br_2}=0,4.0,5=0,2\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

2. \(n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,05\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,05.22,4}{1,4}.100\%=80\%\)

\(\Rightarrow\%V_{CH_4}=100-80=20\%\)

Theo gt ta có: $n_{O_2}=0,6(mol);n_{hh}=0,25(mol)$

a, $CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

Gọi số mol CH4 và C2H4 lần lượt là a;b(mol)

Ta có: $a+b=0,25;2a+3b=0,6\Rightarrow a=0,15;b=0,1$

b, Suy ra $\%V_{CH_4}=60\%;\%V_{C_2H_4}=40\%$

c, Ta có: $n_{CaCO_3}=n_{CO_2}=0,15+0,1.2=0,35(mol)\Rightarrow m_{CaCO_3}=35(g)$

\(a)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ b)\ V_{CH_4} = a(lít) ; V_{C_2H_4} = b(lít)\\ \Rightarrow a + b = 5,6(1)\\ V_{O_2} = 2a + 3b = 13,44(2)\\ (1)(2)\Rightarrow a = 3,36 ; b = 2,24\\ \%V_{CH_4} = \dfrac{3,36}{5,6}.100\% = 60\%\\ \%V_{C_2H_4} = 40\%\\ c) V_{CO_2} = a + 2b = 7,84(lít)\\\)

\(CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ n_{CaCO_3} = n_{CO_2} = \dfrac{7,84}{22,4} = 0,35(mol)\\ \Rightarrow m_{CaCO_3} = 0,35.100 = 35(gam)\)

\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{56}{160}=0,35mol\)

Gọi \(n_{C_2H_4}\) là x \(\Rightarrow V_{C_2H_4}=22,4x\)

       \(n_{C_2H_2}\) là y \(\Rightarrow V_{C_2H_2}=22,4y\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

   x           x                       ( mol )

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

  y          2y                     ( mol )

Ta có:

\(\left\{{}\begin{matrix}22,4x+22,4y=5,6\\x+2y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

\(\Rightarrow V_{C_2H_4}=22,4.0,15=3,36l\)

\(\Rightarrow V_{C_2H_2}=22,4.0,1=2,24l\)

\(\%V_{C_2H_4}=\dfrac{3,36}{5,6}.100=60\%\)

\(\%V_{C_2H_2}=\dfrac{2,24}{5,6}.100=40\%\)

nhh khí = 5,6/22,4 = 0,25 (mol)

Gọi nC2H4 = a (mol); nC2H2 = b (mol)

a + b = 0,25 (1)

nBr2 = 56/160 = 0,35 (mol)

PTHH: 

C2H4 + Br2 -> C2H4Br2

Mol: a ---> a 

C2H2 + 2Br2 -> C2H2Br4

Mol: b ---> 2b

a + 2b = 0,35 (2)

(1)(2) => a = 0,15 (mol); b = 0,1 (mol)

%VC2H2 = 0,15/0,25 = 60%

%VC2H4 = 100% - 60% = 40%

\(a)C_2H_4 + Br_2 \to C_2H_4Br_2\\ b) n_{C_2H_4} = n_{Br_2} = \dfrac{80}{160} = 0,5(mol)\\ \Rightarrow V_{C_2H_4} = 0,5.22,4 = 11,2(lít)\\ V_{CH_4} = 67,2 -11,2 = 56(lít)\\ c) \%V_{CH_4} = \dfrac{56}{67,2}.100\% = 83,33\%\\ \%V_{C_2H_4} = 100\% -83,33\% = 16,67\%\)

1) \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1,68}{6,72}\cdot100\%=25\%\\\%V_{C_2H_2}=75\%\end{matrix}\right.\)

2) Ta có: \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,2}{\dfrac{5,6}{22,4}}\cdot100\%=80\%\) \(\Rightarrow\%V_{CH_{_4}}=20\%\)

huhuu bạn nào giúp tớ với ....

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