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

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

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

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

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

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

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

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

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

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%

a.\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{12,25}{122,5}=0,1mol\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

    2                       2           3       ( mol )

  0,1                                 0,15

\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)

b.\(V_{kk}=V_{O_2}.5=3,36.5=16,8l\)

c.\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

  3        2                   1         ( mol )

0,5   >   0,15                          ( mol )

0,225              0,15                 ( mol )

\(m_{Fe\left(du\right)}=n_{Fe\left(du\right)}.M_{Fe}=\left(0,5-0,225\right).56=15,4g\)

Cập nhật lại câu hỏi đi bạn :v

nKClO3=0,1(mol)

PTHH: 2 KClO3 -to-> 2 KCl +3 O2

0,1_____________0,1______0,15(mol)

a) mKCl=0,1.74,5=7,45(g)

b) V(O2,đktc)=0,15.22,4=3,36(l)

\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

PTHH: \(2KClO_3\rightarrow2KCl+3O_2\)

               0,1             0,1         0,15    (mol)

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

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

\(2KClO_3\rightarrow3O_2+2KCl\)

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

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

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

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