a)

M hóa trị n

n_{H_2}={7,84}/{22,4}=0,35(mol)

M_{hh\ khi}=15,84.2=31,68(g//mol)

->{n_{N_2O}}/{n_{NO}}={31,68-30}/{44-31,68}=3/{22}

Lại có n_{N_2O}+n_{NO}={5,6}/{22,4}=0,25(mol)

->n_{N_2O}=0,03(mol);n_{NO}=0,22(mol)

TN1:

Đặt n_{Mg}=x(mol);n_M=y(mol)

Mg+H_2SO_4->MgSO_4+H_2

2M+2nH_2SO_4->M_2(SO_4)_n+nH_2

Theo PT: n_{Mg}+0,5n.n_M=n_{H_2}=0,35

->x+0,5ny=0,35(1)

m_X=24x+yM=14,8(2)

TN2:

4Mg+10HNO_3->4Mg(NO_3)_2+N_2O+5H_2O

3Mg+8HNO_3->3Mg(NO_3)_2+2NO+4H_2O

8M+10nHNO_3->8M(NO_3)_n+nN_2O+5nH_2O

3R+4nHNO_3->3R(NO_3)_n+nNO+2nH_2O

Theo PT: 3n_{NO}+8n_{N_2O}=2n_{Mg}+n.n_R

->2x+ny=0,9(3)

(1)(2)(3)-> Vô nghiệm.

-> Hóa trị R thay đổi.

TN2:

4Mg+10HNO_3->4Mg(NO_3)_2+N_2O+5H_2O

3Mg+8HNO_3->3Mg(NO_3)_2+2NO+4H_2O

8M+30HNO_3->8M(NO_3)_3+3N_2O+15H_2O

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

Theo PT: 2x+3y=3n_{NO}+8n_{N_2O}=0,9(4)

(1)(4)->2x+3y-2x-ny=0,9-0,7=0,2

->y(3-n)=0,2

->y={0,2}/{3-n}

(4)->2x+{0,6}/{3-n}=0,9

->x=0,45-{0,3}/{3-n}={1,05-0,45n}/{3-n}

(2)->24.{1,05-0,45n}/{3-n}+{0,2M}/{3-n}=14,8

->n=2;M=56 là nghiệm.

->M:\ Fe

b)

y=n_{Fe}={0,2}/{3-2}=0,2(mol)

x=n_{Mg}={14,8-0,2.56}/{24}=0,15(mol)

Theo PT:

n_{Fe(NO_3)_3}=y=0,2(mol)

n_{Mg(NO_3)_2}=x=0,15(mol)

n_{HNO_3}=10n_{N_2O}+4n_{NO}=1,18(mol)

->n_{HNO_3\ du}=1,18.10\%=0,118(mol)

m_{dd\ spu}=14,8+{(1,18+0,118).63}/{15\%}-31,68.0,25=552,04(g)

Trong Z:

C\%_{Fe(NO_3)_3}={0,2.242}/{552,04}.100\%\approx 8,77\%

C\%_{Mg(NO_3)_2}={0,15.148}/{552,04}.100\%\approx 4,02\%

C\%_{HNO_3\ du}={0,118.63}/{552,04}.100\%\approx 1,35\%