Đáp án:
\(a\%=20\%;b\%=8,55\%\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaOH + HN{O_3}\xrightarrow{{}}NaN{O_3} + {H_2}O\)
\(Ba{(OH)_2} + 2HN{O_3}\xrightarrow{{}}Ba{(N{O_3})_2} + 2{H_2}O\)
Ta có:
\({V_{dd{\text{HN}}{{\text{O}}_3}}} = \frac{{224}}{{1,12}} = 200{\text{ ml = 0}}{\text{,2 lít}}\)
\( \to {n_{HN{O_3}}} = 0,2.4,5 = 0,9{\text{ mol}}\)
\({m_{NaOH}} = 150.a\% ;{m_{Ba{{(OH)}_2}}} = 150.b\% \)
\( \to {n_{NaOH}} = \frac{{150a\% }}{{40}} = 3,75a\% \)
\({n_{Ba{{(OH)}_2}}} = \frac{{150b\% }}{{137 + 17.2}} = \frac{{150}}{{171}}b\% \)
Ta có:
\({n_{HN{O_3}}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 3,75a\% + \frac{{150}}{{171}}b\% .2 = 0,9\)
Ta có:
\(b\% = 0,4275a\% \)
\(a\%=20\%;b\%=8,55\%\)
