Đáp án:
\(C{\% _{MgC{l_2}}} = 22,5\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{6}{{24}} = 0,25mol\\
{n_{HCl}} = 2{n_{Mg}} = 0,5mol\\
{m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
{m_{ddHCl}} = \dfrac{{18,25 \times 100}}{{18,25}} = 100g\\
{n_{{H_2}}} = {n_{Mg}} = 0,25mol\\
{m_{ddspu}} = 6 + 100 - 0,25 \times 2 = 105,5g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,25mol\\
{m_{MgC{l_2}}} = 0,25 \times 95 = 23,75g\\
C{\% _{MgC{l_2}}} = \dfrac{{23,75}}{{105,5}} \times 100\% = 22,5\%
\end{array}\)