a, PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

b, \(n_{O_2}=\dfrac{53,76}{22,4}=2,4\left(mol\right)\\ n_{O_2}=2,4.32=76,8\left(g\right)\)

Bảo toàn khối lượng: \(m_{KClO_3}=76,8+168,2=245\left(g\right)\)

c, Theo pthh: \(n_{KClO_3\left(pư\right)}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.2,4=1,6\left(mol\right)\\ \Rightarrow\%m_{KClO_3\left(phân.huỷ\right)}=\dfrac{1,6.122,5}{245}=80\%\)

\(a,PTHH:2KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\\ b,n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{O_2}=0,3\cdot32=9,6\left(g\right)\\ \Rightarrow m_{KMnO_4\left(bđ\right)}=m_{\text{chất rắn}}+m_{O_2}=109,6\left(g\right)\\ c,n_{MnO_2}=0,3\left(mol\right)\\ \Rightarrow m_{MnO_2}=0,3\cdot87=26,1\left(g\right)\\ \Rightarrow\%_{MnO_2}=\dfrac{26,1}{100}\cdot100\%=26,1\%\\ \Rightarrow\%_{KMnO_4}=100\%-26,1\%=73,9\%\)

\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)

=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)

PTHH: 2KClO₃ --t^o,MnO₂--> 2KCl + 3O₂

             0,3825------------------->0,57375

=> \(V_{O_2}=0,57375.22,4=12,852\left(l\right)\)

2KClO₃-to>2KCl+3O₂

0,45---------------------0,675 mol

n _KClO3=\(\dfrac{55,125}{122,5}\)=0,45 mol

=>H=85%

=>V_O2=0,675.22,4.\(\dfrac{85}{100}\)=12,852l

Sửa đề: 1,896 → 0,896

Ta có: \(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,05\left(mol\right)\)

Mà: \(n_{O_2\left(TT\right)}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{0,04}{0,05}.100\%=80\%\)

2KMnO₄ (63/316 mol) \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂\(\downarrow\) + O₂\(\uparrow\) (63/632 mol).

a. Thể tích khí oxi thu được ở đktc là:

V=63/632.22,4=882/395 (lít).

b. Số mol khí oxi phản ứng là (14,4-11,2)/32=0,1 (mol) > 63/632 (mol).

Kết luận: Giả thiết câu b không xảy ra.

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

a) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

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

=> \(m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)

b) \(n_{KClO_3}=\dfrac{490}{122,5}=4\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

                4-------------->4---->6

=> \(m_{KCl}=4.74,5=298\left(g\right)\)

=> \(m_{O_2}=6.32=192\left(g\right)\)

2KClO₃ \(\underrightarrow{t^o}\) 2KCl + 3O₂

a, \(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{KClO_3}=\dfrac{0,3.2}{3}=0,2mol\\ m_{KClO_3}=0,2.122,5=24,5g\)

b, \(n_{KClO_3}=\dfrac{490}{122,5}=4mol\)

\(\Rightarrow m_{KCl}=4.74,5=298g\)

\(n_{O_2}=\dfrac{4.3}{2}=6mol\\ m_{O_2}=6.32=192g\)

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%