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%

\(n_{KClO3}=\dfrac{5,25}{122,5}=\dfrac{3}{70}\left(mol\right)\)

a) PTHH : \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\) 

                     \(\dfrac{3}{70}\)             \(\dfrac{3}{70}\)       \(\dfrac{9}{140}\)

b) \(V_{O2\left(dktc\right)}=\dfrac{9}{140}.22,4=1,44\left(l\right)\)

c) \(m_{KCl\left(lt\right)}=\dfrac{3}{70}.74,5=\dfrac{447}{140}\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{2,235}{\dfrac{447}{140}}.100\%=70\%\)

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

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

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

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

Đổi 672ml = 0,672 lít

Ta có: \(n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: \(2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\)

a. Theo PT: \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,03=0,02\left(mol\right)\)

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

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

=> \(m_{KCl}=0,02.74,5=1,49\left(g\right)\)