Đáp án:
6,11g
Giải thích các bước giải:
\(\begin{array}{l}
{P_1}:\\
M{g^{2 + }} + 2O{H^ - } \to Mg{(OH)_2}\\
N{H_4}^ + + O{H^ - } \to N{H_3} + {H_2}O\\
{n_{Mg{{(OH)}_2}}} = \dfrac{{0,58}}{{58}} = 0,01\,mol \Rightarrow {n_{M{g^{2 + }}}} = 0,01\,mol\\
{n_{N{H_3}}} = \dfrac{{0,672}}{{22,4}} = 0,03\,mol \Rightarrow {n_{N{H_4}^ + }} = 0,03\,mol\\
{P_2}:\\
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
{n_{BaS{O_4}}} = \dfrac{{4,66}}{{233}} = 0,02\,mol \Rightarrow {n_{S{O_4}^{2 - }}} = 0,02\,mol\\
BTDT:{n_{C{l^ - }}} + 2{n_{S{O_4}^{2 - }}} = 2{n_{M{g^{2 + }}}} + {n_{N{H_4}^ + }}\\
\Rightarrow {n_{C{l^ - }}} = 2 \times 0,01 \times 2 + 0,03 \times 2 - 2 \times 0,02 \times 2 = 0,02\,mol\\
{m_E} = 0,02 \times 35,5 + 0,01 \times 2 \times 24 + 0,03 \times 2 \times 18 + 0,02 \times 2 \times 96 = 6,11g
\end{array}\)
