2Zn+O2-to>2ZnO

7\130--7\260

n Zn=\(\dfrac{7}{130}mol\)

=>VO2=\(\dfrac{7}{260}22,4\)=0,6l

b)

2KClO3-to>2KCl+3O2

=>7\390---------------7\260

=>m KClO3=\(\dfrac{7}{390}\).122,5=2,198g

\(n_{Zn}=\dfrac{3,5}{65}=\dfrac{7}{130}mol\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(\dfrac{7}{130}\)    \(\dfrac{7}{260}\)

\(V_{O_2}=\dfrac{7}{360}\cdot22,4=0,6l\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(\dfrac{7}{390}\)                         \(\dfrac{7}{260}\)

\(m_{KClO_3}=\dfrac{7}{390}\cdot122,5=2,2g\)

3,25 g Zn chứ bạn, bạn check lại đề ik :)

2Zn+O2-to>2ZnO

0,05--0,025---0,05

n Zn=0,05 mol

=>=>VO2=0,025.22,4=0,56l

2KClO3-to->2KCl+3O2

1\60------------------------0,025

=>m KClO3=\(\dfrac{1}{60}\).122,5=2,041g

\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

0,05    0,025

\(V_{O_2}=0,025\cdot22,4=0,56l\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(\dfrac{1}{60}\)                           0,025

\(m_{KClO_3}=\dfrac{1}{60}\cdot122,5=2,042g\)

cảm ơn 2 bạn

nC = 1,8 : 12 = 0,15 (mol) 
pthh : C+O2 --> CO2 
       0,15>0,15 (mol) 
=> V O2 = 0,15 .22,4 = 3,36 (l) 
=> Vkk = 3,36 : 1/5 = 16,8 (L)
nZn = 13 : 65 = 0,2 (mol) 
 pthh : 2Zn  + O2 -t-> 2ZnO
           0,2-----> 0,1 (mol) 
=>VO2 = 0,1.22,4 = 2,24 (l) 
=> Vkk = 2,24  : 1/5 = 11,2 (l) 
nAl = 2,7 : 27 = 0,1 (mol) 
pthh : 4Al + 3O2 --t--->2 Al2O3 
          0,1-->0,075 (mol) 
=> VO2 = 0,075 . 22,4 = 1, 68 (l) 
=> VKk = 1,68 :  1/5 = 8,4 (l)
 

a, nC = 1,8/12 = 0,15 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 0,15 ---> 0,3

Vkk = 0,3 . 5 . 22,4 = 33,6 (l)

b, nZn = 13/65 = 0,2 (mol)

PTHH: 2Zn + O2 -> (t°) 2ZnO

Mol: 0,2 ---> 0,1

Vkk = 0,1 . 5 . 22,4 = 11,2 (l)

c, nAl = 2,7/27 = 0,1 (mol)

PTHH: 2Al + 3O2 -> (t°) 2Al2O3

Mol: 0,1 ---> 0,075

Vkk = 0,075 . 5 . 22,4 = 8,4 (l)

a)\(n_C=\dfrac{1,8}{12}=15mol\)

\(C+O_2\underrightarrow{t^o}CO_2\)

15   15     15

\(V_{O_2}=15\cdot22,4=336l\)

\(\Rightarrow V_{kk}=5V_{O_2}=1680l\)

b)\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

0,2      0,1     0,2

\(V_{O_2}=0,1\cdot22,4=2,24l\)

\(V_{kk}=5V_{O_2}=11,2l\)

c)\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

0,1     0,075   0,05

\(V_{O_2}=0,075\cdot22,4=1,68l\)

\(\Rightarrow V_{kk}=1,68\cdot5=8,4l\)

a) \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

           0,4<--0,3<---------0,2

=> m_Al = 0,4.27 = 10,8(g)

b) C1: V_O2 = 0,3.22,4 = 6,72(l)

C2: Theo ĐLBTKL: m_O2 = 20,4 - 10,8 = 9,6(g)

=> \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)=>V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c) V_kk = 6,72 : 20% = 33,6(l)

\(n_C=\dfrac{24}{12}=2\left(mol\right)\)

\(C+O_2\underrightarrow{^{t^0}}CO_2\)

\(2.......2\)

\(V_{kk}=5V_{O_2}=5\cdot2\cdot22.4=224\left(l\right)\)

Oxi chiếm 20% em nhé !

Oxi chiếm 20%. 

nZn = 13/65 = 0.2 (mol) 

2Zn + O2 -to-> 2ZnO 

0.2......0.1

VO2 = 0.1 * 22.4 = 2.24 (l)

Bài 1 :

\(n_{Na}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Na+O_2\rightarrow2Na_2O\)

..0,1....0,025....0,05.......

a, \(V_{O_2}=n.22,4=0,56\left(l\right)\)

b, \(m=m_{Na_2o}=n.M=3,1\left(g\right)\)

Bài 2 :

\(n_{Al}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

..0,1...0,075...

\(\Rightarrow n_{O_2}=0,075\left(mol\right)\)

Mà : \(\Sigma n_{O_2}=\dfrac{V}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow n_{O_2\left(Mg\right)}=0,4-0,075=0,325\left(mol\right)\)

\(2Mg+O_2\rightarrow2MgO\)

.0,65.....0,325........

\(\Rightarrow m_{Mg}=15,6\left(g\right)\)

\(\Rightarrow m_{hh}=2,7+15,6=18,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~14,75\\\%Mg=~85,25\end{matrix}\right.\) %

Bài 3 :

- Gọi số mol Al và Mg lần lượt là x , y

\(4Al+3O_2\rightarrow2Al_2O_3\)

..x....0,75x

\(2Mg+O_2\rightarrow2MgO\)

..y........0,5y...........

Có : \(n_{O_2}=0,75x+0,5y=\dfrac{V}{22,4}=0,1\left(mol\right)\left(I\right)\)

Lại có : \(m_{hh}=m_{Al}+m_{Mg}=27x+24y=3,9\left(II\right)\)

- Giair ( i ) và ( ii ) ta được : \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\) ( mol )

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~69,23\\\%Mg=~30,77\end{matrix}\right.\) %

Vậy ...

Zn + 1/2O2 ---> ZnO

nO2 = 1/2nZnO = 40,5/81/2 = 0,25 mol ---> V = 0,25.22,5 = 5,6 lít.

b) KClO3 ---> KCl + 3/2O2

nKClO3 = 2/3nO2 = 0,5/3 mol ---> m = 122,5.0,5/3 = 20,42 gam.

a)Pt 2Zn + O₂ \(\underrightarrow{t^o}\) 2 ZnO

-ⁿZnO=\(\dfrac{40.5}{81}=0.5\left(mol\right)\)

-Theo pt thì ⁿO₂=\(\dfrac{1}{2}\) ⁿZnO=0.25(mol)

=>^VO₂=0,25.22.4=5.6(l)

b) Pt : 2KClO₃\(\underrightarrow{t^o}\) 2KCl + 3O₂

-Theo phương trình : ⁿKClO₃=\(\dfrac{2}{3}\) ⁿO₂=\(\dfrac{1}{6}\left(mol\right)\)

=>^mKClO₃=1/6.122.5=24.42(g)

Vậy...

\(2C_2H_2+5O_2-^{t^o}\rightarrow4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,125\left(mol\right)\\ \Rightarrow V_{O_2}=2,8\left(l\right)\\ Vìtrongkhôngkhíchứa20\%O_2\\ \Rightarrow V_{kk}=\dfrac{2,8}{20\%}=14\left(l\right)\)