\(-n=\dfrac{m}{M}=\dfrac{47,4}{158}0,3\left(mol\right)\)
\(-PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2\downarrow+O_2\uparrow\)
2 1 1 1
0,3 0,15 0,15 0,15
a) \(V_{O_2}=n.24,79=0,15.24,79=3,7185\left(l\right)\)
\(b)m_{MnO_2}=n.M=0,15.87=13,05\left(g\right).\)