Question

Trong phòng thí nghiệmđểcó 6,72 lít khí oxi (ởđktc) thì cần dùng bao nhiêu gam:
a. KMnO4
b. KClO3

Answer 1

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

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

\(0.6.................................................0.3\)

\(m_{KMnO_4}=0.6\cdot158=94.8\left(g\right)\)

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

\(0.2..........................0.3\)

\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

Answer 2

2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,6-----------------------------------0,3 mol

2KClO₃-to>2KCl +3O₂

0,2---------------------0,3 mol

=>n _O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>m _KmnO4=0,6.158=94,8g

=>m _KClO3=0,2.122,5=24,5g