Đáp án:

\( {C_{M{\text{ NaOH}}}} = 0,4M\)

\( {m_{dd\;{{\text{H}}_2}S{O_4}}} = 98{\text{ gam}}\)

\( C{\% _{N{a_2}S{O_4}}} = 2,335\% \)

\(x\%   = 2,6\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

Ta có:

\({n_{N{a_2}O}} = \frac{{6,2}}{{23.2 + 16}} = 0,1{\text{ mol}}\)

\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,1.2 = 0,2{\text{ mol}}\)

\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,2}}{{0,5}} = 0,4M\)

Trung hòa dung dịch

\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)

Ta có:

\({n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = \frac{{0,2}}{2} = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}}\)

\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{9,8}}{{10\% }} = 98{\text{ gam}}\)

\({m_{dd\;{\text{NaOH}}}} = 500.1,02 = 510{\text{ gam}}\)

\( \to {m_{dd}} = 98 + 510 = 608{\text{ gam}}\)

\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1{\text{ mol}}\)

\( \to {m_{N{a_2}S{O_4}}} = 0,1.(23.2 + 32 + 16.4) = 14,2{\text{ gam}}\)

\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{14,2}}{{608}}.100\%  = 2,335\% \)

Phản ứng xảy ra:

\(N{a_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2NaCl\)

\( \to {n_{BaS{O_4}}} = \frac{{11,65}}{{233}} = 0,05{\text{ mol < }}{{\text{n}}_{N{a_2}S{O_4}}}\)

Vậy \(Na_2SO_4\) dư

\( \to {n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,05{\text{ mol}}\)

\( \to {m_{BaC{l_2}}} = 0,05.(137 + 35,5.2) = 10,4{\text{ gam}}\)

\(x\%  = C{\% _{BaC{l_2}}} = \frac{{10,4}}{{400}}.100\%  = 2,6\% \)

Đáp án:Bài làm mình trong ảnh bạn tham khảo nhé^^

Giải thích các bước giải: