Đáp án:
$ {V_{NO}} = 2,24\,\,l$
Giải thích các bước giải:
Gọi số mol của $Mg$ và $Al$ ở mỗi phần lần lượt là $x$ và $y$ mol.
Xét phần 1:
${n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15\,\,mol$
Phương trình hóa học:
$Mg + 2HCl \to MgC{l_2} + {H_2}$
$\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$2Al + 6HCl \to 2AlC{l_3} + 3{H_2}$
$\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,1,5y\,\,\,\,\,\,\,\,mol$
$\to {n_{{H_2}}} = x + 1,5y = 0,15\,\,\,mol\,→2x + 3y = 0,3 (*)$
Xét phần 2:
Khí không màu, hóa nâu trong không khí là khí $NO$.
Phương trình hóa học:
$3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_3} + 2NO + 4{H_2}O$
$\,\,x \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2}{3}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O$
$\,\,y\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$ \to {n_{NO}} = \frac{{2x}}{3} + y\,\, = \frac{{2x + 3y}}{3} = \frac{{0,3}}{3} = 0,1\,\,mol $
$ \to {V_{NO}} = 0,1.22,4 = 2,24\,\,l$
