Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
1,
\(\begin{array}{l}
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl\\
{n_{{H_2}S{O_4}}} = \dfrac{{180 \times 15}}{{100 \times 98}} = 0,28mol\\
{n_{BaC{l_2}}} = \dfrac{{320 \times 10}}{{100 \times 208}} = 0,15mol\\
\to {n_{{H_2}S{O_4}}} > {n_{BaC{l_2}}}
\end{array}\)
Suy ra có \({H_2}S{O_4}\) dư
Vậy dung dịch sau phản ứng có: \({H_2}S{O_4}\) dư và \(HCl\)
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư= 0,28 - 0,15 = 0,13mol\\
\to {n_{HCl}} = 2{n_{BaC{l_2}}} = 0,3mol\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,15mol\\
{m_{{\rm{dd}}}} = 180 + 320 - (0,15 \times 233) = 465,05g\\
\to C{\% _{{H_2}S{O_4}}}dư= \dfrac{{0,13 \times 98}}{{465,05}} \times 100\% = 2,74\% \\
\to C{\% _{HCl}} = \dfrac{{0,3 \times 36,5}}{{465,05}} \times 100\% = 2,35\%
\end{array}\)
2,
\(\begin{array}{l}
MgC{l_2} + Ba{(OH)_2} \to Mg{(OH)_2} + BaC{l_2}\\
{n_{MgC{l_2}}} = 0,2mol\\
{n_{Ba{{(OH)}_2}}} = 0,225mol\\
\to {n_{Ba{{(OH)}_2}}} > {n_{MgC{l_2}}}
\end{array}\)
Suy ra \(Ba{(OH)_2}\) dư
\(\begin{array}{l}
Mg{(OH)_2} \to MgO + {H_2}O\\
\to {n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,2mol\\
\to {n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,2mol\\
\to {m_{MgO}} = 8g
\end{array}\)
Dung dịch A gồm: \(Ba{(OH)_2}\) dư và \(BaC{l_2}\)
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}}dư= 0,225 - 0,2 = 0,025mol\\
{n_{BaC{l_2}}} = {n_{MgC{l_2}}} = 0,2mol\\
\to C{M_{Ba{{(OH)}_2}}}dư= \dfrac{{0,025}}{{0,25}} = 0,1M\\
\to C{M_{BaC{l_2}}} = \dfrac{{0,2}}{{0,25}} = 0,8M
\end{array}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}} = 1,12 \times 250 = 280g\\
C{\% _{Ba{{(OH)}_2}}}dư= \dfrac{{0,025 \times 171}}{{280}} \times 100\% = 1,53\% \\
C{\% _{BaC{l_2}}} = \dfrac{{0,2 \times 208}}{{280}} \times 100\% = 14,86\%
\end{array}\)
