1, PTHH: 2H2+O2--->2H2O

nH2= \(\dfrac{13,4}{22,4}\)\(\approx\) 0,6 mol

theo pt: nO2= \(\dfrac{1}{2}\). nH2= \(\dfrac{1}{2}\). 0,6= 0,3 mol

=> VO2= 0,3.22,4=6,72 l

3, PTHH: 2H2+O2-->2H2O

nH2= \(\dfrac{11,2}{22,4}\)= 0,5 mol

nO2= \(\dfrac{11,2}{22,4}\)= 0,5 mol

Ta có tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,5}{1}\)

--> H2 pứ hết, O2 dư

Theo pt: nO2_(pứ)= \(\dfrac{1}{2}\). nH2=\(\dfrac{1}{2}.0,5=0,25\) mol

=> mO2= 0,25. 32= 8 g

2, pt: 2H2+O2---> 2H2O

nH2= \(\dfrac{13,44}{22,4}=0,6\) mol

Theo pt: nO2= \(\dfrac{1}{2}\). nH2= \(\dfrac{1}{2}.0,6\)= 0,3 mol

=> VO2= 0,3.22,4= 6,72 l

1, 2H2O---> 2H2+O2

nH2= \(\dfrac{5,6}{22,4}\)= 0,25 mol

Theo pt: nH2O= nH2= 0,25 mol

=> mH2O= 0,25.18= 4,5 g

a, 4Al+3O2---> 2Al2O3

Al2O3+3H2SO4---> Al2(SO4)3+3H2O

Al2(SO4)3+6NaOH---> 2Al(OH)3+3Na2SO4

2Al(OH)3--t^o--> Al2O3+3H2O

b, 4Fe+3O2--->2Fe2O3

Fe2O3+3H2SO4---> Fe2(SO4)3+3H2O

Fe2(SO4)3+6NaOH--->2Fe(OH)3+3Na2SO4

2Fe(OH)3--t^o--> Fe2O3+3H2O

Fe2O3+6HCl---> 2FeCl3+3H2O

2P+5Cl2---> 2PCl5

4Fe+3O2---> 2Fe2O3

(x-3)(x+3)=0

<=> \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

a, CuSO4+ Ba(OH)2----> Cu(OH)2+ BaSO4

b, CuCl2+Fe----> FeCl2+Cu

c, H2SO4+Ba(OH)2----> BaSO4+2H2O

d, BaCl2+Na2SO4----> BaSO4+2NaCl

e, CuSO4+2NaOH----> Cu(OH)2+Na2SO4