a)
nFe=\(\frac{7}{56}\)=0,125(mol)
nZn=\(\frac{6,5}{65}\)=0,1(mol)
Ta có PT
3Fe + 2O2 ---> Fe3O4
0,125..0,833.....0,04167
2Zn + O2 ---> 2ZnO
0,1.....0,05........0,1
mchất rắn= 232.0,04167 + 81.0,1=17,77(g)
b)n\(O_2\)= 0,833 + 0,05 = 0,1333(mol)
=> V\(O_2\)=1,333.22,4=3(l)
=> Vkk=3.5=15(l)