Theo PTHH :

\( n_{H_2} = n_{H_2O} = \dfrac{3,6}{18} = 0,2(mol)\\ \Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\)

a) 

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

             0,4-->0,2------->0,4

=> \(m_{H_2O}=0,4.18=7,2\left(g\right)\)

b) V_O2 = 0,2.22,4 = 4,48 (l)

c)

C1: m_O2 = 0,2.32 = 6,4 (g)

C2:

Theo ĐLBTKL: \(m_{O_2}=m_{H_2O}-m_{H_2}=7,2-0,4.2=6,4\left(g\right)\)

\(m_{CO}=2,4.28=67,2g\)

\(n_{N_2}=\dfrac{8,96}{22,4}=0,4mol\)      \(\Rightarrow m_{N_2}=0,4.28=11,2g\)

\(m_{hh}=67,2+11,2+16=94,4g\)

\(V_{CO}=2,4.22,4=53,76l\)

\(V_{SO_2}=\dfrac{16}{44}.22,4=8,1l\)

\(\Rightarrow V_{hh}=53,76+8,1+8,96=70,82l\)

`a)PTHH:`

`C_2 H_4 + 3O_2` $\xrightarrow{t^o}$ `2CO_2 + 2H_2 O`

 `1/60`        `0,05`                `1/30`                    `(mol)`

`b)n_[C_2 H_4]=[11,2]/[22,4]=0,5(mol)`

   `n_[O_2]=[1,12]/[22,4]=0,05(mol)`

Có: `[0,5]/1 < [0,05]/3 =>C_2 H_4` dư

  `=>V_[C_2 H_4(dư)]=1/60 .22,4=0,37(l)`

`c)V_[CO_2]=1/30 .22,4=0,75(l)`

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.1..............................................0.05\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\ PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \left(mol\right).......0,1\rightarrow.......................0,05\\ V_{O_2}=0,05.22,4=1,12\left(l\right)\)

Gọi CTHH của khí X là $RO_3$

Ta có:  

\(M_{RO_3}=R+48=\dfrac{16}{\dfrac{4,4}{44}.2}=80\Rightarrow R=32\left(S\right)\)

Vậy CTHH khí X là $SO_3$

\(a) 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ b)V_{O_2} = \dfrac{1}{2}V_{H_2} = 0,7(lít)\\ V_{không\ khí} = \dfrac{V_{O_2}}{20\%} = \dfrac{0,7}{20\%} = 3,5(lít)\\ c) n_{H_2O} = n_{H_2} = \dfrac{1,4}{22,4} = 0,0625(mol)\\ m_{H_2O} = 0,0625.18 = 1,125(gam)\)