n_{HNO_3}=0,5.1=0,5(mol)

n_{N_2O}={1,008}/{22,4}=0,045(mol)

10H^{+}+2NO_3^{-}+8e->N_2O+5H_2O

10H^{+}+NO_3^{-}+8e->NH_4^{+}+3H_2O

Do 10n_{N_2O}=0,45<0,5=n_{HNO_3}-> Tạo NH_4NO_3

Theo bán pu: n_{HNO_3}=10n_{N_2O}+10n_{NH_4NO_3}

->n_{NH_4NO_3}={0,5-0,045.10}/{10}=0,005(mol)

Đặt n_{Mg}=x(mol);n_{Zn}=y(mol)

->24x+65y=8,9(1)

Bảo toàn e: 2x+2y=8n_{N_2O}+8n_{NH_4NO_3}=0,4(2)

(1)(2)->x=y=0,1(mol)

Bảo toàn Mg,Zn: n_{Mg(NO_3)_2}=x=0,1(mol);n_{Zn(NO_3)_2}=y=0,1(mol)

->m=0,1.148+0,1.189+0,005.80=34,1(g)