Question

Answer 1

Ta có: \(n_{KMnO_2}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

\(pthh:2KMnO_4\overset{t^o}{--->}K_2MnO_4+MnO_2+O_2\uparrow\)

a. Theo pt: \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,05.22,4=1,12\left(lít\right)\)

b. Theo pt: \(n_{MnO_2}=n_{O_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{MnO_2}=0,05.87=4,35\left(g\right)\)