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

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PTHH: \(n_{KClO_3}=\dfrac{0,25.2}{3}\approx0,17\left(mol\right)\)

Vậy muốn điều chế 5,6 lít O₂ cần dùng số gam Kali clorat:

\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,17.122,5=20,825g\)

             \(n_{O2}\)=\(\dfrac{V}{22,4}\)=\(\dfrac{5,6}{22,4}\)=0,25 (mol)

     PT : 2KClO3 →^to  2KCl + 3O2

số mol:      \(\dfrac{1}{6}\)      ←    \(\dfrac{1}{6}\)      ← 0,25

⇒ m_KClO3 = n . M = \(\dfrac{1}{6}\) . 122,5 ∼_∼ 20,41(g)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5=\dfrac{245}{12}\left(g\right)\)

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

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

Theo PTHH : 

\(n_{KClO_3\ phản\ ứng} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow n_{KClO_3\ cần\ dùng} = \dfrac{0,2}{70\%} = \dfrac{2}{7}(mol)\\ \Rightarrow m_{KClO_3\ cần\ dùng} = \dfrac{2}{7}.122,5 = 35(gam)\)

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

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

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

nO₂ = \(\dfrac{17,92}{22,4}\)= 0,8 mol

2KMnO₄   t^o→ K₂MnO₄  + O₂  + MnO₂

Theo phương trình phản ứng, để thu được 0,8 mol hay 17,92 lít O₂ thì cần 0,8.2 = 1,6 mol KMnO₄

=> mKMnO₄ = 1,6.(39+55+16.4) = 252,8 gam

a) 2KClO₃ (7/75 mol) \(\underrightarrow{t^o}\) 2KCl (7/75 mol) + 3O₂\(\uparrow\) (0,14 mol).

b) Số mol khí oxi là 4,48/32=0,14 (mol).

Khối lượng kali clorat cần dùng là 7/75.122,5=343/30 (g).

Khối lượng chất rắn thu được là 7/75.74,5=1043/150 (g).

\(a,PTHH:2KClO_3\underrightarrow{t^o,MnO_2}2KCl+3O_2\uparrow\\ b,n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ Theo.pt:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{KClO_3}=\dfrac{2}{15}.122,5=\dfrac{49}{3}\left(g\right)\)

\(a.\)

\(n_{KClO_3}=\dfrac{3.675}{122.5}=0.03\left(mol\right)\)

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

\(0.03........................0.045\)

\(V_{O_2}=0.045\cdot22.4=1.008\left(l\right)\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(\Rightarrow n_{KClO_3}=\dfrac{0.5\cdot2}{3}=\dfrac{1}{3}mol\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)

Chọn B