Đáp án:
\(\begin{array}{l}
a)\\
D:Magie(Mg)\\
b)\\
{V_{{H_2}}} = 0,448l\\
C{\% _{MgS{O_4}}} = 1,19\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2D + 2xHCl \to 2DC{l_x} + x{H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_D} = \dfrac{2}{x}{n_{{H_2}}} = \dfrac{{0,4}}{x}mol\\
{M_D} = 4,8:\dfrac{{0,4}}{x} = 12x\\
x = 2 \Rightarrow {M_D} = 24\\
\Rightarrow D:Magie(Mg)\\
b)\\
{n_D} = \dfrac{1}{{10}} \times 0,2 = 0,02mol\\
{m_D} = 0,02 \times 24 = 0,48g\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = {n_{Mg}} = 0,02mol\\
{V_{{H_2}}} = 0,02 \times 22,4 = 0,448l\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,02mol\\
{m_{MgS{O_4}}} = 0,02 \times 120 = 2,4g\\
{m_{ddspu}} = 0,48 + 200 - 0,02 \times 2 = 200,44g\\
C{\% _{MgS{O_4}}} = \dfrac{{2,4}}{{200,44}} \times 100\% = 1,19\%
\end{array}\)
