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

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

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

đề thiếu

Ét ô Ét 

Xét hợp chất: R_y(SO₄)_x

Ta có:\(\dfrac{2R}{96x}=\dfrac{20}{80}\)

=>R=12x (1)

Xét hợp chất R_yO_x:

Ta có:%R= \(\dfrac{2R}{2R+16x}.100\%=\dfrac{R}{R+8x},100\%\left(2\right)\)

Thay (1) vào (2) ta có :%R= \(\dfrac{12x}{12x+8x}.100\%=60\%\)

đó là phần a hay b vậy bạn

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

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

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

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

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

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

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

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

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

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

C

đáp án là C nha bạn

Câu 1:

nKClO3=12,25122,5=0,1mol

2KClO3→(to,MnO2)2KCl+3O2

0,1                                                0,15 ( mol )

VO2=0,15.22,4=3,36l

=> Chọn B

\(C_mH_nONa+\left(m+\dfrac{n}{2}-\dfrac{1}{4}\right)O_2\underrightarrow{t^o}\left(m-\dfrac{1}{2}\right)CO_2+\dfrac{n}{2}H_2O+\dfrac{1}{2}Na_2CO_3\)

n_Zn = \(\frac{m}{M}\) = \(\frac{13}{65}\)= 0,2 (mol)

a) PTHH: Zn + H₂SO₄ \(\rightarrow\)ZnSO₄ + H₂

                0,2 -> 0,2       ->     0,2   -> 0,2

b) V_H2 = n . 22,4 = 0,2 . 22,4 = 4,48 (l)

c) m_ZnSO4 = n . M = 0,2 . 161 = 32,2 (g)

d) n_Fe3O4 = \(\frac{m}{M}\) = \(\frac{23,2}{232}\)= 0,1 (mol)

PTHH: 4H₂ + Fe₃O₄ \(\rightarrow\)3Fe + 4H₂O

                         0,1     ->    0,3

m_Fe = n . M = 0,3 . 56 = 16,8 (g)