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 : 

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

\(a.\)

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

\(b.\)

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

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

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

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

\(c.\)

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

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

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

\(\Rightarrow a=0.2\)

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

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

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

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

\(n_{O_2}=\dfrac{43.2}{32}=1.35\left(mol\right)\)

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

\(0.9...........................1.35\)

\(H\%=\dfrac{0.9}{1}\cdot100\%=90\%\)

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

______1_____________1,5 (mol)

⇒ m_{O2 (lí thuyết)} = 1,5.32 = 48 (g)

Mà: m_{O2 (thực tế)} = 43,2 (g)

\(\Rightarrow H\%=\dfrac{43,2}{48}.100\%=90\%\)

Bạn tham khảo nhé!

\(n_{O_2}\)=\(\dfrac{43,2}{32}=1,35\left(mol\right)\)

PTHH 2KClO₃-----t^o--->2KCl +3O₂

=>\(n_{O_2\left(lt\right)}=1.\dfrac{3}{2}=1,5\left(mol\right)\)

=>H%=\(\dfrac{1,35}{1,5}.100\%=90\%\)

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

\(a,PTHH:2KClO_3\rightarrow\left(^{t^o}_{MnO_2}\right)2KCl+3O_2\\ b,m_{KClO_3}=m_{KCl}+m_{O_2}\\ c,m_{KCl}=m_{KClO_3}-m_{O_2}=14,9\left(g\right)\\ d,\text{Số phân tử }O_2:\text{Số phân tử }KCl=3:2\\ \text{Số phân tử }O_2:\text{Số phân tử }KClO_3=3:2\)

\(a,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{395}{158}=2,5(mol)\\ \Rightarrow n_{O_2}=1,25(mol)\\ \Rightarrow V_{O_2}=1,25.22,4=28(l)\\ \Rightarrow V_{O_2(tt)}=28.85\%=23,8(l)\)

\(b,n_{O_2}=\dfrac{67,2}{22,4}=3(mol)\\ 2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=6(mol)\\ \Rightarrow m_{KMnO_4}=6.158=948(g)\\ \Rightarrow m_{KMnO_4(tt)}=\dfrac{948}{80\%}=1185(g)\)

\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{67,2}{22,4}=3\left(mol\right)\\ Theo.PTHH:n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.3=2\left(mol\right)\\ m_{KClO_3}=n.M=2.122,5=245\left(g\right)\)

PTHH:

\(2KClO_3\rightarrow2KCl+3O_2\)

BTKL:

\(m_{KClO_3}=m_{KCl}+m_{O_2}\)

\(\Leftrightarrow m_{KCl}=m_{KClO_3}-m_{O_2}=24,5-9,6=14,9g\)

PTHH: \(2KClO_3\rightarrow2KCl+3O_2\)

\(1\rightarrow1,5\left(mol\right)\)

Theo phương trình: \(n_{O_2lt}=\dfrac{1.3}{2}=1,5\left(mol\right)\)

Khối lượng \(O_2\) thu được theo lý thuyết là :

\(m_{O_2lt}=1,5.32=48\left(g\right)\)

Hiệu suất phản ứng là:

\(H=\dfrac{43,2}{44}.100\%=90\%\)

n O2(tt) = 43,2/32 = 1,35(mol)

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$

Theo PTHH : 

n O2(lt) = 3/2 n KClO3 = 1,5(mol)

H = n O2(tt) / n O2(lt) .100% = 1,35/1,5   .100% = 90%

\(2KClO_3\rightarrow3O_2+2KCl\)

\(m_{KClO_3}=m_{O_2}+m_{KCl}\)

\(\Rightarrow m_{KCl}=m_{KClO_3}-m_{KCl}=24,5-9,6=14,9\left(g\right)\)