Đáp án:
\(\begin{array}{l}
5)\\
\% {m_{Mg}} = 51,61\% \\
\% {m_{Al}} = 48,39\% \\
6)\\
\% {m_{Mg}} = 74,71\% \\
\% {m_{Al}} = 25,29\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5)\\
Al \to A{l^{ + 3}} + 3e\\
Mg \to M{g^{ + 2}} + 2e\\
{N^{ + 5}} + 3e \to NO\\
2{N^{ + 5}} + 8e \to {N_2}O\\
hh:Mg(a\,mol),Al(b\,mol)\\
24a + 27b = 5,58\\
2a + 3b = 0,1 \times 3 + 0,03 \times 8\\
\Rightarrow a = 0,12;b = 0,1\\
\% {m_{Mg}} = \dfrac{{0,12 \times 24}}{{5,58}} \times 100\% = 51,61\% \\
\% {m_{Al}} = 100 - 51,61 = 48,39\% \\
6)\\
Zn \to Z{n^{ + 2}} + 2e\\
Mg \to M{g^{ + 2}} + 2e\\
{N^{ + 5}} + 3e \to NO\\
2{N^{ + 5}} + 10e \to {N_2}\\
hh:Mg(a\,mol),Zn(b\,mol)\\
24a + 65b = 3,855\\
2a + 2b = 0,04 \times 3 + 0,015 \times 10\\
\Rightarrow a = 0,12;b = 0,015\\
\% {m_{Mg}} = \dfrac{{0,12 \times 24}}{{3,855}} \times 100\% = 74,71\% \\
\% {m_{Al}} = 100 - 74,71 = 25,29\%
\end{array}\)
