Đáp án:

Giải thích các bước giải:

1)

M_{hh}=19.2=38(g//mol)

->{n_{N_2}}/{n_{N_2O}}={44-38}/{38-28}=3/5

Mà n_{N_2}+n_{N_2O}={0,896}/{22,4}=0,04

->n_{N_2}=0,015(mol);n_{N_2O}=0,025(mol)

2)

M->M^{+n}+n e

2N^{+5}+10e->N_2^0

2N^{+5}+8e->N_2^{+1}

Bảo toàn e: n.n_M=10n_{N_2}+8n_{N_2O}=0,35(mol)

->n_M={0,35}/n(mol)

->M_M={4,2}/{{0,35}/n}=12n

->n=2;M_M=24

->M:\ Mg

3)

n_{Mg}={0,35}/2=0,175(mol)

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

5Mg+12HNO_3->5Mg(NO_3)_2+N_2+6H_2O

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

Theo PT:

n_{HNO_3\ pu}=12n_{N_2}+10n_{N_2O}=0,43(mol)

n_{Mg(NO_3)_2}=n_{Mg}=0,175(mol)

->n_{HNO_3\ du}=0,5-0,43=0,07(mol)

n_{Na}={6,9}/{23}=0,3(mol)

2Na+2HCl->2NaCl+H_2

2Na+2H_2O->2NaOH+H_2

NaOH+HNO_3->NaNO_3+H_2O

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

->n_{Mg(OH)_2}={5,8}/{58}=0,1(mol)

Theo PT:

\sum n_{NaOH}=2n_{Mg(OH)_2}+n_{HNO_3\ du}=0,27(mol)

n_{H_2}=1/2n_{Na}=0,15(mol)

->V=0,15.22,4=3,36(l)

Lại có n_{NaOH}+n_{HCl}=n_{Na}

->n_{HCl}=0,3-0,27=0,03(mol)

200ml=0,2l

->x={0,03}/{0,2}=0,15M