P->P^{+5}+5e
S->S^{+6}+6e
N^{+5}+1e->N^{+4}
Bảo toàn e: n_{NO_2}=5a+6b
2H^{+}+NO_3^{-}+1e->NO_2+H_2O
Theo bán pu: n_{HNO_3\ pu}=2n_{NO_2}=10a+12b(mol)
->n_{HNO_3\ du}=20\%(10a+12b)=2a+2,4b(mol)
NaOH+HNO_3->NaNO_3+H_2O
->n_{NaOH}=n_{HNO_3\ du}=2a+2,4b(mol)
