Đáp án:
2,96g
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,02\,mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,02\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,05\,mol\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
{n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,02\,mol\\
{n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = 0,02\,mol\\
m = 0,02 \times 58 + 0,02 \times 90 = 2,96g
\end{array}\)