Đáp án:
\(\begin{array}{l}
a)\\
{m_{Mg}} = 3,6g\\
{m_{MgO}} = 4g\\
b)\\
{m_{BaS{O_4}}} = 69,9g
\end{array}\)
\({m_{Mg{{(OH)}_2}}} = 14,5g\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
{m_{MgO}} = 7,6 - 3,6 = 4g\\
b)\\
{n_{{H_2}S{O_4}}} = 0,3 \times 1 = 0,3mol\\
{n_{MgO}} = \dfrac{4}{{40}} = 0,1mol\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - {n_{Mg}} - {n_{MgO}} = 0,05mol\\
MgS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Mg{(OH)_2}\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{MgS{O_4}}} = {n_{Mg}} + {n_{MgO}} = 0,25mol\\
{n_{BaS{O_4}}} = {n_{MgS{O_4}}} + {n_{{H_2}S{O_4}d}} = 0,25 + 0,05 = 0,3mol\\
{m_{BaS{O_4}}} = 0,3 \times 233 = 69,9g\\
{n_{Mg{{(OH)}_2}}} = {n_{MgS{O_4}}} = 0,25mol\\
{m_{Mg{{(OH)}_2}}} = 0,25 \times 58 = 14,5g
\end{array}\)
