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)

$C_2H_4 + Br_2 \to C_2H_4Br_2$

$n_{C_2H_4} = n_{Br_2} = \dfrac{24}{160} = 0,15(mol)$

$n_X = \dfrac[7,84}{22,4} = 0,35(mol)$

$\Rightarrow n_{CH_4} = 0,35 - 0,15 = 0,2(mol)$

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O$
$n_{O_2} = 2n_{CH_4} + 3n_{C_2H_4} = 0,85(mol)$
$V_{O_2} = 0,85.22,4 = 19,04(lít)$
$V_{không\ khí} = V_{O_2} : 20\% = 95,2(lít)$

b)

$M_X = \dfrac{0,2.16 + 0,15.28}{0,35} = 21,14(g/mol)$
$d_{X/không\ khí} = \dfrac{21,14}{29} = 0,73$

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) \(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: C₂H₂ + 2Br₂ --> C₂H₂Br₄

            0,1<-----0,2

=> m_C2H2 = 0,1.26 =2,6 (g)

\(\%m_{C_2H_2}=\dfrac{2,6}{8}.100\%=32,5\%\)

\(\%m_{CH_4}=\dfrac{8-2,6}{8}.100\%=67,5\%\)

b) \(n_{CH_4}=\dfrac{8-2,6}{16}=0,3375\left(mol\right)\)
PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

        0,3375->0,675 

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

              0,1---->0,25

=> V_O2 = (0,675 + 0,25).22,4 = 20,72 (l)

=> V_kk = 20,72.5 = 103,6 (l)

refer

Gọi x, y lần lượt là số mol của C₂H₄, C₂H₂ ( x, y > 0 )

n_Br2 = 0,2 mol

C₂H₄ + Br₂ → C₂H₄Br₂

x............x...............x

C₂H₂ + 2Br₂ → C₂H₂Br₄

y.............2y..............y

Ta có hệ

{28x+26y=4,1x+2y=0,2{28x+26y=4,1x+2y=0,2

⇒ {x=0,1y=0,05{x=0,1y=0,05

⇒ %_C2H4 =   31,7%

C₂H₄ + 3O₂ ---t^o---> 2CO₂ + 2H₂O

0,1.........0,3

⇒ VO2 = 0,3.22,4 = 6,72 (l)

2C₂H₂ + 5O₂ ---t^o---> 4CO₂ + 2H₂O

0,05.......0,125

⇒ VO2 = 0,125.22,4 = 2,8 (l)

⇒ _VO2 = 6,72 + 2,8 = 9,52 (l)

a.\(n_{Br_2}=\dfrac{32}{160}=0,2mol\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

  0,1       0,2                          ( mol )

\(m_{C_2H_2}=0,1.26=2,6g\)

\(\%m_{C_2H_2}=\dfrac{2,6}{8}.100=32,5\%\)

\(\%m_{CH_4}=100\%-32,5\%=67,5\%\)

b.\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

      0,1       0,25                                     ( mol )

\(n_{CH_4}=\dfrac{8-2,6}{16}=0,3375mol\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,3375  0,675                                 ( mol )

\(V_{kk}=\left(0,25+0,675\right).22,4.5=103,6l\)              

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

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

\(\Rightarrow x+y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(1\right)\)

\(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y=\dfrac{24}{160}=0,15\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x=y=0,05\left(mol\right)\)

\(\Rightarrow\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,05.22,4}{2,24}.100\%=50\%\)