Đáp án:
\( C{\% _{HN{O_3}}} = 3,84\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(HCl + AgN{O_3}\xrightarrow{{}}AgCl + HN{O_3}\)
Ta có:
\({n_{HCl}} = \frac{{1,792}}{{22,4}} = 0,08{\text{ mol}}\)
\({m_{dd{\text{ AgN}}{{\text{O}}_3}}} = 50.1,19 = 59,5{\text{ gam}}\)
\( \to {m_{AgN{O_3}}} = 59,5.10\% = 5,95{\text{ gam}}\)
\( \to {n_{AgN{O_3}}} = \frac{{5,95}}{{108 + 62}} = 0,035{\text{ mol < }}{{\text{n}}_{HCl}}\)
Vậy \(HCl\) dư.
\( \to {n_{AgCl}} = {n_{HN{O_3}}} = {n_{AgN{O_3}}} = 0,035{\text{ mol}}\)
\( \to {m_{AgCl}} = 0,035.143,5 = 5,0225{\text{ gam}}\)
\({m_{HN{O_3}}} = 0,035.63 = 2,205{\text{ gam}}\)
BTKL:
\({m_{HCl}} + {m_{dd{\text{ AgN}}{{\text{O}}_3}}} = {m_{dd}} + {m_{AgCl}}\)
\( \to 0,08.36,5 + 59,5 = {m_{dd}} + 5,0225\)
\( \to {m_{dd}} = 57,3975{\text{ gam}}\)
\( \to C{\% _{HN{O_3}}} = \frac{{2,205}}{{57,3975}} = 3,84\% \)
