Đáp án:
\(\begin{array}{l}
a)\\
\% Al = 23,48\% \\
\% Mg = 20,87\% \\
\% Cu = 55,65\% \\
b)\\
C{\% _{HN{O_3}}} = 50,4\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 4HN{O_3} \to Mg{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
a)\\
{n_{N{O_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{m_{Al}} = 2,7g\\
hh:Al,Mg(a\,mol),Cu(b\,mol)\\
2a + 2b = 0,4(1)\\
24a + 64b = 11,5 - 2,7 = 8,8(2)\\
\text{Từ (1) và (2)} \Rightarrow a = 0,1;b = 0,1\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
\% Al = \dfrac{{2,7}}{{11,5}} \times 100\% = 23,48\% \\
\% Mg = \dfrac{{2,4}}{{11,5}} \times 100\% = 20,87\% \\
\% Cu = \dfrac{{6,4}}{{11,5}} \times 100\% = 55,65\% \\
b)\\
{n_{HN{O_3}}} = 4 \times {n_{Mg}} + 4 \times {n_{Cu}} = 0,4 + 0,4 = 0,8mol\\
{m_{HN{O_3}}} = 0,8 \times 63 = 50,4g\\
{m_{{\rm{dd}}HN{O_3}}} = 100 \times 1 = 100g\\
C{\% _{HN{O_3}}} = \dfrac{{50,4}}{{100}} \times 100\% = 50,4\%
\end{array}\)
