n_{HNO_3}=0,8.1=0,8(mol)

M_Z=18.2=36(g//mol)

->{n_{N_2}}/{n_{N_2O}}={44-36}/{36-28}=1

->n_{N_2}=n_{N_2O}=0,5.{1,12}/{22,4}=0,025(mol)

Lại có n_{HNO_3}>10n_{N_2O}+12n_{N_2}

-> Tạo NH_4NO_3

->n_{NH_4NO_3}={n_{HNO_3}-10n_{N_2O}-12n_{N_2}}/{10}=0,025(mol)

-> Muối gồm Mg(NO_3)_2:x(mol);Fe(NO_3)_3:y(mol)

->148x+242y=52-0,025.80=50(1)

Bảo toàn Fe,Mg,electron: 2x+3y=10n_{N_2}+8n_{N_2O}+8n_{NH_4NO_3}

->2x+3y=0,65(2)

(1)(2)->x=0,1825;y=0,095

NH_4NO_3+NaOH->NaNO_3+NH_3+H_2O

Mg(NO_3)_2+2NaOH->Mg(OH)_2+2NaNO_3

Fe(NO_3)_3+3NaOH->Fe(OH)_3+3NaNO_3

->n_{Mg(OH)_2}=0,1825(mol);n_{Fe(OH)_3}=0,095(mol)

->m=0,1825.58+0,095.107=20,75(g)

\sum n_{NaOH}=n_{NH_4NO_3}+2x+3y=0,675(mol)