Question

1. Phân hủy hoàn toàn 1,58g muối kali pemanganat ( KMnO₄ )

A. Viết PTHH

B. Tính thể tích khí oxi (đktc ) thu đc trong pứ trên

Answer 1

Answer 2

\(n_{KMnO_4}=\dfrac{1,58}{158}=0,01\left(mol\right)\)
a. PTHH: \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\uparrow\)
b. Theo PT ta có: \(n_{O_2}=\dfrac{0,01.1}{2}=0,005\left(mol\right)\)
=> \(V_{O_2}=0,005.22,4=0,112\left(l\right)\)

Answer 3

a) \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\)

b) \(n_{KMnO_4}=\dfrac{1,58}{158}=0,01\left(mol\right)\)

PTHH: \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo PTHH: \(n_{KMnO_4}:n_{O_2}=2:1\)

\(\Rightarrow n_{O_2}=n_{KMnO_4}.\dfrac{1}{2}=0,01.\dfrac{1}{2}=\dfrac{1}{200}\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=\dfrac{1}{200}.22,4=0,112\left(l\right)\)