n H2O=\(\dfrac{3,6}{18}\)=0,2 mol
2H2O-đp->2H2+O2
0,2-------------------0,1 mol
2O2+3Fe-to>Fe3O4
0,1----------------0,05 mol
=>x=VO2=0,1.22,4=2,24l
=>y=m Fe3O4=0,05.232=11,6g
\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\\a,PTHH:2H_2O\rightarrow\left(đp\right)2H_2+O_2\\ b,n_{O_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow x=V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,2O_2+3Fe\rightarrow\left(t^o\right)Fe_3O_4\\ \Rightarrow n_{Fe_3O_4}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow y=m_{sp}=0,05.232=11,6\left(g\right)\)
Chắc đề chỗ g là chữ y em ha