a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KCl}=\dfrac{1,49}{74,5}=0,02\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,03\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,03.24,79=0,7437\left(l\right)\)

b, Theo PT: \(n_{KClO_3\left(TT\right)}=n_{KCl}=0,02\left(mol\right)\)

\(\Rightarrow m_{KClO_3\left(TT\right)}=0,02.122,5=2,45\left(g\right)\)

\(\Rightarrow H=\dfrac{2,45}{3,5}.100\%=70\%\)

2KClO3=>2KCl+3O2

a, nKCl=1,49/74,5=0,02(mol)

=>nO2=0,03(mol)

=>V O2=0,03.22,4=0,672(l)

b, nKClO3=0,02(mol)

mKClO3=0,02.122,5=2,45(g)

H(KClO3)=2,45/3,5.100%=70%

\(n_{KClO_3}=\dfrac{7}{122,5}=\dfrac{2}{35}\left(mol\right)\)

PTHH :

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

2/35        2/35         3/35

\(V_{O_2}=\dfrac{3}{35}.24,79\approx2,1249\left(l\right)\)

\(m_{KCl}=\dfrac{2}{35}.74,5\approx4,257\left(g\right)\)

\(H=\dfrac{2,98}{4,257}.100\%=70\%\)

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

b, \(n_{KCl}=\dfrac{0,745}{74,5}=0,01\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,015\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,015.24,79=0,37185\left(l\right)\)

\(m_{O_2}=0,015.32=0,48\left(g\right)\)

c, \(n_{KClO_3\left(pư\right)}=n_{KCl}=0,01\left(mol\right)\)

\(\Rightarrow m_{KClO_3\left(pư\right)}=0,01.122,5=1,225\left(g\right)\)

\(\Rightarrow H=\dfrac{1,225}{2,5}.100\%=49\%\)

Gọi hiệu suất phản ứng là a

$2SO_2 + O_2 \xrightarrow{t^o,V_2O_5} 2SO_3$

Ta thấy : 

V SO2 / 2 = V O2 nên hiệu suất tính theo số mol của SO2 hoặc O2

V SO2 phản ứng = 4,48a(lít)

V O2 phản ứng = 2,24a(lít)

V SO3 = V SO2 pư = 4,48a(lít)

Sau phản ứng, khí gồm :

 SO2 : 4,48 - 4,48a(lít)

O2 : 2,24 - 2,24a(lít)

SO3 : 4,48a(lít)

Suy ra : 

4,48 - 4,48a + 2,24 - 2,24a + 4,48a = 5,6

=> a = 0,5 = 50%

Hỗn hợp khí gồm : 

SO2 : 4,48 -4,48.0,5 = 2,24 lít

O2 : 2,24 - 2,24.0,5 = 1,12 lít

SO3 : 4,48.0,5 = 2,24 lít