Question

nung 15,8 gam kalipermanganat thu được bao nhiêu lít oxi(đktc) biết H =90%

cảm ơn mn

Answer 1

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

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

\(\Rightarrow n_{O_2\left(thực\right)}=\dfrac{1}{2}\cdot0,1\cdot90\%=0,045\left(mol\right)\) \(\Rightarrow V_{O_2}=0,045\cdot22,4=1,008\left(l\right)\)

Answer 2

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

PTHH: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

Mol:          0,1                                                0,05

⇒ \(V_{O_2}=90\%.0,05.22,4=1,008\left(l\right)\)