Hòa tan hoàn toàn 27,04 gam hỗn

Chọn đáp án A

Tính được: $n_{N_{2}O}=\mathrm{0,12}mol$ và $n_{H_2}=\mathrm{0,16}mol$

Dung dịch Y chứa Mg²⁺, Al³⁺ (x mol), NH₄⁺ (y mol), Na⁺ (z mol) và SO₄^2- (1,08 mol)

Theo đề:$\left\{\begin{array}{l}2n_{M{g}^{2+}}+4x+y=\mathrm{2,28}\\ 40n_{Mg{\left(OH\right)}_2}=40n_{M{g}^{2+}}=\mathrm{27,84}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{M{g}^{2+}}=\mathrm{0,48}mol\\ 4x+y=\mathrm{1,32}\left(1\right)\end{array}\right.$và $\stackrel{BTDT}{\to }3x+y+z=\mathrm{1,2}\left(2\right)$

$\stackrel{BT:H}{\to }{n}_{H_{2}O}=\mathrm{0,92}-2y\stackrel{BTKL}{\to }-27x+18y+62z=\mathrm{4,48}\left(3\right)$

Từ (1), (2), (3) suy ra: x = 0,32 ; y = 0,04 ; z = 0,2

$\stackrel{BT:N}{\to }{n}_{Mg{\left(N{O}_3\right)}_2}=\mathrm{0,04}mol\Rightarrow n_{Mg}=\mathrm{0,44}mol$

Ta có: $\left\{\begin{array}{l}{n}_{Al}+2n_{A{l}_2{O}_3}=\mathrm{0,32}\\ 27n_{Al}+102n_{A{l}_2{O}_3}=\mathrm{10,56}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{Al}=\mathrm{0,24}mol\\ n_{A{l}_2{O}_3}=\mathrm{0,04}mol\end{array}\right.\Rightarrow \%m_{Al}=\mathrm{23,96}\%$