Question

Nung 32,67g KClO3 có 25% tạp chất(có xúc tác MnO2).Tính thể tích O2(đktc) thoát ra

Answer 1

\(m_{KClO_3}=\left(100\%-25\%\right).32,67=24,5025\left(g\right)\)
=> \(n_{KClO_3}=\dfrac{24,5025}{122,5}=0,20\left(mol\right)\)
PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)
Theo PT ta có: \(n_{O_2}=\dfrac{0,20.3}{2}=0,3\left(mol\right)\)
=> \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

Answer 2

^mKClO₃ = 32,67.(100% - 25%) = 24,5025 (g)

ⁿKClO₃ = \(\dfrac{24,5025}{122,5}\) \(\approx\) 0,2 (mol)

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

Theo PT: ⁿO₂ = \(\dfrac{3}{2}\)ⁿKClO₃ = \(\dfrac{3}{2}.0,2\) = 0,3 (mol)

=> VO₂ = 0,3.22,4 = 6,81 (l)

Vậy thể tích oxi thoát ra là 6,81 lít

Answer 3

\(m_{KClO_3}=32,67.\left(100\%-25\%\right)=24,5025\left(g\right)\)

\(n_{KClO_3}=\dfrac{24,5025}{122,5}=0,20\left(mol\right)\)

PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^0}2KCl+3O_2\)

Theo PTHH: \(n_{KClO_3}:n_{O_2}=2:3\)

\(\Rightarrow n_{O_2}=n_{KClO_3}.\dfrac{3}{2}=0,20.\dfrac{3}{2}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)