Đáp án:
1,\(C{M_{NaOH}}dư= \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M\) và \(C{M_{NaCl}}= \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M\)
2,\(C{M_{{H_2}S{O_4}}}dư= \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M\)
3,
\(C{\% _{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{0,1 \times 98}}{{96,3}} \times 100\% = 10,18\% \)
\(C{M_{{H_2}S{O_4}}}dư= \dfrac{{0,1}}{{0,2}} = 0,5M\)
Giải thích các bước giải:
1,
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}} = 0,1mol\\
{n_{NaOH}} = 0,2mol\\
\to {n_{NaOH}} > {n_{HCl}}
\end{array}\)
Suy ra dung dịch sau phản ứng có NaCl và NaOH dư
\(\begin{array}{l}
\to {n_{NaOH}}dư= 0,2 - 0,1 = 0,1mol\\
\to {n_{NaCl}} = {n_{HCl}} = 0,1mol\\
\to C{M_{NaOH}}dư= \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M\\
\to C{M_{NaCl}} = \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M
\end{array}\)
2,
\(\begin{array}{l}
{H_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} > {n_{Ca{{(OH)}_2}}}
\end{array}\)
Suy ra dung dịch sau phản ứng có \({H_2}S{O_4}\) dư
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư= 0,2 - 0,1 = 0,1mol\\
\to C{M_{{H_2}S{O_4}}}dư= \dfrac{{0,1}}{{0,3}} = \dfrac{1}{3}M
\end{array}\)
3,
\(\begin{array}{l}
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{100 \times 17,1}}{{100 \times 171}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} > {n_{Ba{{(OH)}_2}}}
\end{array}\)
Suy ra dung dịch sau phản ứng có \({H_2}S{O_4}\) dư
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư= 0,2 - 0,1 = 0,1mol\\
\to {n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,1mol\\
{m_{{\rm{dd}}}} = {m_{{H_2}S{O_4}}} + {m_{Ba{{(OH)}_2}}}dd - {m_{BaS{O_4}}} = 0,2 \times 98 + 100 - 0,1 \times 233 = 96,3g\\
\to C{\% _{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{0,1 \times 98}}{{96,3}} \times 100\% = 10,18\% \\
\to C{M_{{H_2}S{O_4}}}dư= \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)
