Đáp án:

\({C_{M{\text{ (N}}{{\text{H}}_4}{)_2}S{O_4}}} = {C_{M{\text{ N}}{{\text{H}}_4}N{O_3}}} = 1M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(N{H_4}^ +  + O{H^ - }\xrightarrow{{}}N{H_3} + {H_2}O\)

\(B{a^{2 + }} + S{O_4}^{2 - }\xrightarrow{{}}BaS{O_4}\)

Ta có: 

\({n_{N{H_3}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{N{H_4}^ + }};{n_{BaS{O_4}}} = \frac{{23,3}}{{233}} = 0,1{\text{ mol = }}{{\text{n}}_{S{O_4}^{2 - }}}\)

Bảo toàn điện tích:

\({n_{N{H_4}^ + }} = {n_{N{O_3}^ - }} + 2{n_{S{O_4}^{2 - }}} \to {n_{N{O_3}^ - }} = 0,3 - 0,1.2 = 0,1{\text{ mol}}\)

\( \to {n_{N{H_4}N{O_3}}} = {n_{N{O_3}^ - }} = 0,1{\text{ mol;}}{{\text{n}}_{{{(N{H_4})}_2}S{O_4}}} = {n_{S{O_4}^{1 - }}} = 0,4{\text{ mol}}\)

\( \to {C_{M{\text{ (N}}{{\text{H}}_4}{)_2}S{O_4}}} = {C_{M{\text{ N}}{{\text{H}}_4}N{O_3}}} = \frac{{0,1}}{{0,1}} = 1M\)

$Ba(OH)_2\to Ba^{2+}+ 2OH^-$

$n_{BaSO_4\downarrow}= 0,1 mol$

$n_{NH_3\uparrow }= 0,3 mol$

$Ba^{2+} + SO_4^{2-}\to BaSO_4$

$NH_4^+ + OH^-\to NH_3+H_2O$

$\Rightarrow n_{NH_4^+}=0,3 mol; n_{SO_4^{2-}}= 0,1 mol$

$n_{(NH_4)_2SO_4}= 0,1 mol$

$\Rightarrow n_{NH_4^+\text{trong muối sunfat}}= 0,1.2=0,2 mol$

$\Rightarrow n_{NH_4NO_3}= n_{NH_4^+\text{trong muối nitrat}}= 0,3-0,2=0,1 mol$

$C_{M_{(NH_4)_2SO_4}}=\dfrac{0,1}{0,1}=1M$

$C_{M_{NH_4NO_3}}=\dfrac{0,1}{0,1}=0,1M$