Đáp án:
a, \({V_{{H_2}}} = 2,24l\)
b,
\(C{\% _{HCl dư }} = \dfrac{{7,7}}{{152,3}} \times 100\% = 5,06\% \)
\(C{\% _{MgC{l_2}}} = \dfrac{{9,5}}{{152,3}} \times 100\% = 6,24\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,1mol\\
{m_{HCl}} = \dfrac{{150 \times 10}}{{100}} = 15g \to {n_{HCl}} = 0,4mol\\
{n_{Mg}} < \dfrac{{{n_{HCl}}}}{2}\\
{n_{{H_2}}} = {n_{Mg}} = 0,1mol \to {V_{{H_2}}} = 2,24l\\
{m_{HCl dư }} = 15 - (0,2 \times 36,5) = 7,7g\\
{m_{MgC{l_2}}} = 0,1 \times 95 = 9,5g\\
{m_{{\rm{dd}}}} = 2,5 + 150 - 0,1 \times 2 = 152,3\\
C{\% _{HCl dư }} = \dfrac{{7,7}}{{152,3}} \times 100\% = 5,06\% \\
C{\% _{MgC{l_2}}} = \dfrac{{9,5}}{{152,3}} \times 100\% = 6,24\%
\end{array}\)
