Đáp án:
Phản ứng xảy ra: $2KMnO_4\to K_2MnO_4+MnO_2+O_2\qquad(1)$
$3Fe+2O_2\to Fe_3O_4\qquad(2)$
$n_{Fe}=\dfrac{16,8}{56}=0,3(mol)$
Theo PTHH $(2)\to n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac 23.0,3=0,2(mol)$
Theo PTHH $(1)\to n_{KMnO_4}=2.n_{O_2}=2.0,2=0,4(mol)$
$\to m_{KMnO_4}=0,4.158=63,2(g)$