Question

Answer 1

\(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right);n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ 2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ a,n_{O_2}=\dfrac{n_{KMnO_4}}{2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,n_{Fe}=\dfrac{3}{2}.n_{O_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow m_{Fe}=0,075.56=4,2\left(g\right)\)