Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 0,896l\\
b)\\
C{\% _{MgC{l_2}}} = 24,55\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
a)\\
{n_{Mg}} = \dfrac{m}{M} = \dfrac{{0,96}}{{24}} = 0,04mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,04mol\\
{V_{{H_2}}} = n \times 22,4 = 0,04 \times 22,4 = 0,896l\\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,04mol\\
{m_{MgC{l_2}}} = n \times M = 0,04 \times 95 = 3,8g\\
{m_{ddspu}} = 0,96 + 14,6 - 0,04 \times 2 = 15,48g\\
C{\% _{MgC{l_2}}} = \dfrac{{3,8}}{{15,48}} \times 100\% = 24,55\%
\end{array}\)