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

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

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

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

$2KClO_3  \xrightarrow{t^o,MnO_2} 2KCl + 3O_2$
$n_{KCl} = \dfrac{17,88}{74,5} = 0,24(mol)$
$n_{KClO_3\ pư} = 0,24 : 80\% = 0,3(mol)$
$m = 0,3.122,5 = 36,75(gam)$

   Theo định luật bảo toàn khối lượng, ta có khối lượng khí oxi thu được là:

    m O 2  = 24,5 – 13,45 = 11,05(g)

   Khối lượng thực tế oxi thu được:  m O 2  = (11,05 x 80)/100 = 8,84 (g)

2Mg+O₂-to>2MgO

0,15---0,075 mol

2KClO₃-to->2KCl+3O₂

0,05-----------------------0,075

n _Mg=\(\dfrac{3,6}{24}\)=0,15 mol

=>V_O2=0,075.22,4=1,68l

=>m _KClO3=0,05.122,5=6,125g

Sao lại cho Mg vào mà ''khối lượng kali clorat'' là sao em

 a. \(n_{Mg}=\dfrac{3.6}{24}=0,15\left(mol\right)\)

PTHH : 2Mg + O₂ ----t^o----> 2MgO

             0,15   0,075             

\(V_{O_2}=0,075.22,4=1,68\left(l\right)\)

b. PTHH : 2KClO₃ \(\xrightarrow[MnO_2]{t^o}\) 2KCl + 3O₂

                   0,05                         0,075

\(m_{KClO_3}=0,05.122,5=6,125\left(g\right)\)

\(n_{KCl}=\dfrac{17.88}{74.5}=0.24\left(mol\right)\)

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

\(0.24.............0.24\)

\(m_{KClO_3}=\dfrac{0.24\cdot122.5}{80\%}=36.75\left(g\right)\)

Bài 1:

a, \(S+O_2\underrightarrow{t^o}SO_2\)

b, Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=n_S=0,1\left(mol\right)\Rightarrow m_{SO_2}=0,1.64=6,4\left(g\right)\)

c, \(n_{O_2}=n_S=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

Bài 2:

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

b, Bạn xem lại đề nhé, pư không tạo thành MnO₂.

Bài 3:

a, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

c, \(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\Rightarrow V_{H_2O}=0,1.22,4=2,24\left(l\right)\)

d, \(n_{CuO}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)