a)

4Fe+3O_2  $$\xrightarrow{t^o}$$  2Fe_2O_3

Fe+4HNO_3->Fe(NO_3)_3+NO+2H_2O

Fe_2O_3+6HNO_3->2Fe(NO_3)_3+3H_2O

b)

n_{NO}={0,448}/{22,4}=0,02(mol)

Theo PT: n_{Fe\ du}=n_{NO}=0,02(mol)

->n_{Fe_2O_3}={4,32-0,02.56}/{160}=0,02(mol)

Theo PT: n_{Fe\ pu}=2n_{Fe_2O_3}=0,04(mol)

->m=0,04.56+0,02.56=3,36(g)

b)

Theo PT: n_{HNO_3}=6n_{Fe_2O_3}+4n_{Fe\ du}=0,2(mol)

->V_{dd\ HNO_3}={0,2}/2=0,1(L)=100(mL)