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

b)

\(n_{KClO_3} = \dfrac{36,75}{122,5} = 0,3(mol)\)

Theo PTHH : 

\(n_{KCl} = n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{KCl} = 0,3.74,5 = 22,35(gam)\\ \Rightarrow m_{O_2} = m_{KClO_3} - m_{KCl} = 14,4(gam)\)

c)

Bảo toàn khối lượng : 

\(m_{O_2} = 25 - 15,4 = 9,6(gam)\\ \Rightarrow n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{KClO_3} = 0,2.122,5 = 24,5(gam)\\ \%m_{tạp\ chất}= \dfrac{25-24,5}{25}.100\% = 2\%\)

\(a.\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(b.\)

\(n_{KClO_3}=\dfrac{36.75}{122.5}=0.3\left(mol\right)\)

\(\Rightarrow n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}\cdot0.3=0.45\left(mol\right)\)

\(m_{O_2}=0.45\cdot32=14.4\left(g\right)\)

\(m_{KCl}=0.3\cdot74.5=22.35\left(g\right)\)

\(c.\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(a.............a\)

\(m_{Cr}=m_{KCl}+m_{tc}=25-122.5a+74.5a=15.4\left(g\right)\)

\(\Rightarrow a=0.2\)

\(m_{O_2}=\dfrac{3}{2}\cdot0.2\cdot32=9.6\left(g\right)\)

\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

\(m_{tc}=25-24.5=0.5\left(g\right)\)

\(\%m_{Tc}=\dfrac{0.5}{25}\cdot100\%-2\%\)

2KClO₃-to\xt->2KCl+3O₂

0,1------------------0,1

n KClO3=\(\dfrac{12,25}{122,5}\)=0,1 mol

=>m KCl=0,1.74,5=7,45g

H=\(\dfrac{6,8}{7,45}.100\)=91,275%

b)

2KClO₃-to\xt->2KCl+3O₂

0,2-------------------------0,3 mol

n O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

H=85%

=>m KClO3=0,2.122,5.\(\dfrac{100}{85}\)=28,82g

c)

2KClO₃-to\xt->2KCl+3O₂

0,2------------------------0,3

n KClO3=\(\dfrac{24,5}{122,5}\)=0,2 mol

H=80%

=>m O2=0,3.32.\(\dfrac{80}{100}\)=10,4g

a) 2KClO₃ --t^o--> 2KCl + 3O₂

b) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

               0,1<----------0,1<---0,15

=> \(m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

c) \(m_{KCl}=0,1.74,5=7,45\left(g\right)\)

2KClO3 ---> 2KCl +3O2

nKClo3 = 24,5/122,5 = 0,2 mol

nKCl = nKClo3 =0,2 mol

m Kcl = 0,2 x 74,5 = 14,9g

no2 = 0,2x3:2 = 0,3mol

Vo2 = n.22,4 = 6,72 lít

\(V_{O_2}=n_{O_2}.22,4=0,33.22,4=7,392\left(l\right)\)

2KClO3 -- > 2KCl + O2 

nKClO3 = 73,5 / 122,5 = 0,6 (mol)

mKCl = 0,6 . 74,5 = 44,7 (g)

VO2 = 0,3 . 22,4 = 6,72 (l)

2KClO3--->2KCl+302

nKClO3=122,5:122,5=1 mol

nO2=43,2:32=1,35 mol

So sánh nKClO3/2     >  nO2      --->nO2 hết KClO3 dư  tính theo O2

                 (=0,5)             (=0,45)

Theo pt -->nKClO3 thực =nO2=1,35 

mKClO3=1,35.122,5=165,375 g

H=(165,375/122.5).100%=135%

Mình nhầm bạn nhá phải là nKClO3=nO2.2/3=1,35.2/3=0,9 mol

g KClO3 tt=0,9 mol

H=(0,9/1).100%=90%

\(n_{O_2}=\frac{43,2}{32}=1,35mol\)

PTHH: \(2KClO_3\rightarrow^{t^o}2KCl+3O_2\uparrow\)

Theo phương trình \(n_{KClO_3\left(p/ứ\right)}=\frac{2}{3}n_{O_2}=0,9mol\)

\(\rightarrow H=\frac{0,9.122,5}{122,5}.100\%=90\%\)