\(m_{C_2H_4}=5.6\left(g\right)\)

\(n_{C_2H_4}=\dfrac{5.6}{28}=0.2\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(V_{CH_4}=11.2-0.2\cdot22.4=6.72\left(l\right)\)

\(n_{CH_4}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(m_{CH_4}=0.3\cdot16=4.8\left(g\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^0}2CO_2+2H_2O\)

\(V_{O_2}=\left(0.3\cdot2+0.2\cdot3\right)\cdot22.4=26.88\left(l\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

PTHH:

\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

b--------------->0,5b

\(2Mg+O_2\xrightarrow[]{t^o}2MgO\)

a---------------->a

\(\rightarrow\left\{{}\begin{matrix}24a+27b=5,1\\40a+102b=9,1\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\left(TM\right)\)\

PTHH:

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

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

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)+3H_2\uparrow\)

0,1----------------------------------->0,15

=> V = (0,1 + 0,15).22,4 = 5,6 (l)

a) C2H4 + Br2 ---> C2H4Br2

Khí thoát ra là CH4 (metan) ---> mCH4 = 0,2.16 = 3,2g ---> mC2H4 = 6 - 3,2 = 2,8g. (0,1 mol).

---> %CH4 = 0,2/0,3 = 66,7%; %C2H4 = 33,3%.

b) nCO2 = nCH4 + 2nC2H4 = 0,4 mol.

CO2 + Ca(OH)2 ---> CaCO3 + H2O

CO2 + CaCO3 + H2O ---> Ca(HCO3)2

---> nCaCO3 = 0,2 mol ---> m = 0,2.100 = 20g.

nCa(OH)2=0.3=>nCaCO3=0.3 chứ

Giả sử các khí đo ở điều kiện sao cho 1 mol khí chiếm thể tích V lít

Gọi số mol CH₄, C₂H₂ là a, b (mol)

=> \(a+b=\dfrac{0,05}{V}\) (1)

\(n_{O_2}=\dfrac{0,11}{V}\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

               a-->2a---------->a

             2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

                b--->2,5b-------->2b

=> \(2a+2,5b=\dfrac{0,11}{V}\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{0,03}{V}\left(mol\right)\\b=\dfrac{0,02}{V}\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{\dfrac{0,03}{V}.V}{0,05}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{\dfrac{0,02}{V}.V}{0,05}.100\%=40\%\end{matrix}\right.\)

b) \(n_{CO_2}=a+2b=\dfrac{0,07}{V}\left(mol\right)\)

=> \(V_{CO_2}=\dfrac{0,07}{V}.V=0,07\left(l\right)\)

Tham khảo:

a)

{VC2H2=x(ml)

VC2H4=y(ml)

⇒ x + y = 50(1)

C2H2+5/2O2(to)→2CO2+H2O

C2H4+3O2(to)→2CO2+2H2O

Theo PTHH : 2,5x + 3y = 140(2)

Từ (1)(2) suy ra: x = 20 ; y = 30

Vậy :

%VC2H2=2050.100%=40%%

VC2H4=100%−40%=60%

b)VCO2=2VC2H2+2VC2H4=2.50=100(ml)

a)

{VC2H2=x(ml)

VC2H4=y(ml)

⇒ x + y = 50(1)

C2H2+5/2O2(to)→2CO2+H2O

C2H4+3O2(to)→2CO2+2H2O

Theo PTHH : 2,5x + 3y = 140(2)

Từ (1)(2) suy ra: x = 20 ; y = 30

Vậy :

%VC2H2=2050.100%=40%%

VC2H4=100%−40%=60%

b)VCO2=2VC2H2+2VC2H4=2.50=100(ml)

a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

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

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

b, \(m_{CH_4}=0,05.16=0,8\left(g\right)\)

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

Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)

a) CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

b) \(n_{CH_4}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             0,6-->1,2--------->0,6

=> V_CO2 = 0,6.22,4 = 13,44 (l)

c) V_O2 = 1,2.22,4 = 26,88 (l)

=> V_kk = 26,88.5 = 134,4 (l)

\(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

\(n_{hh}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(V_{CH_4}=2.24\left(l\right)\)

\(n_{CH_4}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(\Rightarrow n_{C_2H_4}=0.15-0.1=0.05\left(mol\right)\)

\(\%CH_4=\dfrac{0.1}{0.15}\cdot100\%=66.67\%\)

\(\%C_2H_4=33.33\%\)

\(CH_4+2O_2\underrightarrow{^{t^0}}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{^{t^0}}2CO_2+2H_2O\)

\(n_{O_2}=0.1\cdot2+0.05\cdot3=0.35\left(mol\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.35\cdot22.4=39.2\left(l\right)\)