Đáp án:
\(\begin{array}{l}
a)\\
m = 19g\\
b)\\
{m_{{\rm{dd}}HN{O_3}}} = 200g\\
c)\\
{V_{{H_2}}} = 2,8l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_2} + 2NO + 4{H_2}O\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
{n_{NO}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:Mg(a\,mol),Cu(b\,mol)\\
\dfrac{2}{3}a + \dfrac{2}{3}b = 0,25(1)\\
148a + 188b = 65,5(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,125;b = 0,25\\
{m_{Mg}} = 0,125 \times 24 = 3g\\
{m_{Cu}} = 0,25 \times 64 = 16g\\
m = 3 + 16 = 19g\\
b)\\
{n_{HN{O_3}}} = \dfrac{8}{3}{n_{Mg}} + \dfrac{8}{3}{n_{Cu}} = 1mol\\
{m_{HN{O_3}}} = 1 \times 63 = 63g\\
{m_{{\rm{dd}}HN{O_3}}} = \dfrac{{63 \times 100}}{{31,5}} = 200g\\
c)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = {n_{Mg}} = 0,125mol\\
{V_{{H_2}}} = 0,125 \times 22,4 = 2,8l
\end{array}\)