Đáp án:
a) Nhôm
b) 8,4l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2R + 2nHCl \to 2RC{l_n} + n{H_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2R + nC{l_2} \to 2RC{l_n}\\
hh:Fe(a\,mol),R(4a\,mol)\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
\Rightarrow a + 4a \times \dfrac{n}{2} = 0,35 \Leftrightarrow a + 2an = 0,35(1)\\
{n_{C{l_2}}} = \frac{{8,4}}{{22,4}} = 0,375\,mol\\
\Rightarrow 1,5a + 4a \times \dfrac{n}{2} = 0,375 \Leftrightarrow 1,5a + 2an = 0,375(2)\\
(2) - (1) \Leftrightarrow 1,5a - a = 0,375 - 0,35 \Leftrightarrow a = 0,05\,\\
{m_R} = 8,2 - 0,05 \times 56 = 5,4g\\
{M_R} = \dfrac{{5,4}}{{0,05 \times 4}} = 27g/mol \Rightarrow R:Al\\
b)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = 0,05 \times \dfrac{3}{2} + 0,05 \times 4 \times \dfrac{3}{2} = 0,375\,mol\\
{V_{S{O_2}}} = 0,375 \times 22,4 = 8,4l
\end{array}\)
