Đáp án + Giải thích các bước giải:

$n_{Fe(NO_3)_3}=\frac{14,52}{242}=0,06(mol)$

$4Fe(NO_3)_3\xrightarrow{t^o}2Fe_2O_3+12NO_2+3O_2$

Theo PT: $n_{NO_2}=3.0,06=0,18(mol);n_{O_2}=\frac{3}{4}.0,06=0,045(mol)$

$\to V_X=V_{NO_2}+V_{O_2}=22,4(0,18+0,045)=5,04(l)$

$\to \overline{M}_X=\frac{0,18.46+0,045.32}{0,18+0,045}=43,2(g/mol)$

$\to d_{X/H_2}=\frac{43,2}{2}=21,6$

Em tham khảo nha :

\(\begin{array}{l}
4Fe{(N{O_3})_3} \to 2F{e_2}{O_3} + 12N{O_2} + 3{O_2}\\
{n_{Fe{{(N{O_3})}_3}}} = \dfrac{{14,52}}{{242}} = 0,06mol\\
{n_{N{O_2}}} = 3{n_{Fe{{(N{O_3})}_3}}} = 0,18mol\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Fe{{(N{O_3})}_3}}} = 0,045mol\\
{n_{hh}} = 0,18 + 0,045 = 0,225mol\\
{V_{hh}} = 0,225 \times 22,4 = 5,04l\\
{M_{hh}} = \dfrac{{0,18 \times 46 + 0,045 \times 32}}{{0,225}} = 43,2dvC\\
{d_{hh/{H_2}}} = \dfrac{{43,2}}{2} = 21,6
\end{array}\)