Đáp án:
\(\begin{array}{l}
V = 0,4l\\
m = 43,9g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{MgO}} = \dfrac{{16}}{{40}} = 0,4mol\\
{n_{HCl}} = 0,82 \times V(mol)\\
{n_{{H_2}S{O_4}}} = 0,59 \times V(mol)\\
\dfrac{{{n_{HCl}}}}{2} + {n_{{H_2}S{O_4}}} = {n_{MgO}}\\
\dfrac{{0,82 \times V}}{2} + 0,59 \times V = 0,4\\
\Rightarrow V = 0,4l\\
{n_{HCl}} = 0,82 \times 0,4 = 0,328mol\\
{n_{MgC{l_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,164mol\\
{m_{MgC{l_2}}} = 0,164 \times 95 = 15,58g\\
{n_{{H_2}S{O_4}}} = 0,59 \times 0,4 = 0,236mol\\
{n_{MgS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,236mol\\
{m_{MgS{O_4}}} = 0,236 \times 120 = 28,32g\\
m = {m_{MgS{O_4}}} + {m_{MgC{l_2}}} = 28,32 + 15,58 = 43,9g
\end{array}\)
