Đáp án:
\(\begin{array}{l}
b)\\
{m_{Mg}} = 2,4g\\
{m_{MgO}} = 2g\\
c)\\
{m_{{\rm{dd}}HCl}} = 150g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
{m_{MgO}} = 4,4 - 2,4 = 2g\\
c)\\
{n_{MgO}} = \dfrac{2}{{40}} = 0,05\,mol\\
{n_{HCl}} = 0,05 \times 2 + 0,1 \times 2 = 0,3\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,3 \times 36,5}}{{7,3\% }} = 150g
\end{array}\)