\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\left(1\right)\\ Fe+2H_2SO_{4\left(đặc\right)}\rightarrow FeSO_4+SO_2\uparrow+2H_2O\left(2\right)\\ 5SO_2+2H_2O+2KMnO_4\rightarrow2H_2SO_4+MnSO_4+K_2SO_4\left(3\right)\)

\(Theo.pt\left(1\right):n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ Theo.pt\left(2\right):n_{Fe}=n_{SO_2}=0,2\left(mol\right)\\ Theo.pt\left(3\right):n_{KMnO_4}=\dfrac{2}{5}n_{SO_2}=\dfrac{2}{5}.0,2=0,08\left(mol\right)\)

\(\Rightarrow V=V_{SO_2}=0,2.22,4=4,48\left(l\right)\\ \Rightarrow V_{ddKMnO_4}=\dfrac{0,08}{2}=0,04\left(l\right)=40\left(ml\right)\)

đề thiếu

Ét ô Ét 

bài 2

Ta sục Br2

Chất làm mất màu C2H2

Chất ko hiện tg :CO2

C2H2+2Br2->C2H2Br4

Bài 3

CaC2+2H2O->Ca(OH)2+C2H2

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

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

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

\(\Rightarrow n_{Br_2}=0,05\left(mol\right);n_{C_2H_4}=0,05\left(mol\right)\)

\(V_{Br_2}=0,05.22,4=1,12\left(l\right)\)

\(V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)

\(\%V_{C_2H_4}=\frac{1,12}{4,48}.100=25\%\)

\(\%V_{CH_4}=100\%-25\%=75\%\)

\(a,n_{C_2H_4Br_2}=\dfrac{9,4}{188}=0,05\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:0,05\leftarrow0,05\leftarrow0,05\\ V_{Br_2}=\dfrac{0,05}{0,1}=0,5\left(l\right)=500\left(ml\right)\)

\(b,\%V_{C_2H_4}=\dfrac{0,05.22,4}{4,48}=25\%\\ \%V_{CH_4}=100\%-25\%=75\%\)

Đặt X có dạng CxHy

Theo bài ra, ta có: %C%H%C%H = 85,7100−85,785,7100−85,7

→ 12xy12xy= 85,714,385,714,3

→ y = 2x            

=> CTHH là C2H4 (thỏa mãn)  

=> Vậy ta chọn C. C2H4

Đặt \(X\)có dạng \(C_xH_y\)

Theo bài ra, ta có: \(\dfrac{\%C}{\%H}\) = \(\dfrac{85,7}{100-85,7}\)

 \(\Rightarrow\frac{12x}{y}\)\(=\)\(\dfrac{85,7}{14,3}\)

\(\Rightarrow y=2x\)         

\(\Rightarrow C_2H_4\)

Đáp án C 

\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Gọi số mol C₂H₄, C₂H₂ là a, b (mol)

=> \(a+b=\dfrac{6,72}{22,4}-0,1=0,2\left(mol\right)\) (1)

\(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

               a--->a

            C₂H₂ + 2Br₂ --> C₂H₂Br₄

              b---->2b

=> a + 2b = 0,3 (2)

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

\(\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,1.28+0,1.26}.100\%=22,857\%\)

\(\%m_{C_2H_4}=\dfrac{0,1.28}{0,1.16+0,1.28+0,1.26}.100\%=40\%\)

\(\%m_{C_2H_2}=\dfrac{0,1.26}{0,1.16+0,1.28+0,1.26}.100\%=37,143\%\)

PTHH : \(C_2H_4\rightarrow2H_2O+2CO_2\)

\(n_{C_2H_4}=\frac{14}{28}=0,5\left(mol\right)\)

Theo phương trình : 

\(n_{O_2}=3n_{C_2H_4}=3.0,5=1,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

\(n_{CO_2}=2n_{C_2H_4}=2.0,5=1\left(mol\right)\)

\(\Rightarrow m_{CO_2}=1.44=44\left(gam\right)\)

PTHH: C2H4 (0,5 mol) + 3O2 (1,5 mol) to→ 2CO2 (1 mol) + 2H2O.

Thể tích khí oxi cần dùng ở đktc: Vkhí oxi=1,5x22,4=33,6 (lít).

Khối lượng khí CO2 sinh ra là: mkhí cacbonic=1x44=44 (gam).

Chọn A.

nhh khí = 7,84/22,4 = 0,35 (mol)

Gọi nCH4 = a (mol); nC2H6 = b (mol)

a + b = 0,35 (1)

nCaCO3 = 50/100 = 0,5 (mol)

PTHH: 

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

0,5 <--- 0,5 <--- 0,5

CH4 + 2O2 -> (t°) CO2 + 2H2O

a ---> 2a ---> a

2C2H6 + 7O2 -> (t°) 4CO2 + 6H2O

b ---> 3,5b ---> 2b

=> a + 2b = 0,5 (2)

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

mCH4 = 0,2 . 16 = 3,2 (g)

mC2H6 = 0,15 . 30 = 4,5 (g)

%mCH4 = 3,2/(3,2 + 4,5) = 41,55%

%mC2H6 = 100% - 41,55% = 58,45%