Đáp án:

\(\begin{array}{l} a,\ C_{M_{NaOH}}=1\ M.\\ b,\ V_{\text{dd H$_2$SO$_4$}}=107,456\ ml.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:Na_2O+H_2O\to 2NaOH\\ n_{Na_2O}=\dfrac{15,5}{62}=0,25\ mol.\\ Theo\ pt:\ n_{NaOH}=2n_{Na_2O}=0,5\ mol.\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\ M.\\ b,\\ PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ Theo\ pt:\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,25\times 98}{20\%}=122,5\ g.\\ \Rightarrow V_{\text{dd H$_2$SO$_4$}}=\dfrac{122,5}{1,14}=107,456\ ml.\end{array}\)

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Đáp án:

\({V_{dd{{\text{H}}_2}S{O_4}}} = 107,456{\text{ ml}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

Ta có:

\({n_{N{a_2}O}} = \frac{{15,5}}{{23.2 + 16}} = 0,25{\text{ mol}}\)

\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,25.2 = 0,5{\text{ mol}}\)

\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,5}}{{0,5}} = 1M\)

Cho dung dịch \(H_2SO_4\) 20% trung hòa dung dịch trên

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

Ta có:

\({n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,25{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,25.98 = 24,5{\text{ gam}}\)

\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{24,5}}{{20\% }} = 122,5{\text{ gam}}\)

\( \to {V_{dd{{\text{H}}_2}S{O_4}}} = \frac{{122,5}}{{1,14}} = 107,456{\text{ ml}}\)