Đáp án:
\({m_{AgCl}} = 43,05{\text{ gam}}\)
\( C{\% _{HN{O_3}}} = 3,386\% \)
\(C{\% _{HCl}} = 1,3\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(AgN{O_3} + HCl\xrightarrow{{}}AgCl + HN{O_3}\)
Ta có:
\({m_{AgN{O_3}}} = 510.10\% = 51{\text{ gam}}\)
\({m_{HCl}} = 91,25.20\% = 18,25{\text{ gam }}\)
\( \to {n_{AgN{O_3}}} = \frac{{51}}{{108 + 62}} = 0,3{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{18,25}}{{36,5}} = 0,5{\text{ mol}}\)
Vì \({n_{HCl}} > {n_{AgN{O_3}}}\) nên \(HCl\) dư
\( \to {n_{AgCl}} = {n_{HN{O_3}}} = {n_{AgN{O_3}}} = 0,3{\text{ mol}}\)
\( \to {n_{HCl{\text{ dư}}}} = 0,5 - 0,3 = 0,2{\text{ mol}}\)
\({m_{AgCl}} = 0,3(108 + 35,5) = 43,05{\text{ gam}}\)
BTKL:
\({m_{dd\;{\text{AgN}}{{\text{O}}_3}}} + {m_{dd\;{\text{HCl}}}} = {m_{dd}} + {m_{AgCl}}\)
\( \to 510 + 91,25 = {m_{dd}} + 0,3.(108 + 35,5)\)
\( \to {m_{dd}} = 558,2{\text{ gam}}\)
\({m_{HN{O_3}}} = 0,3.63 = 18,9{\text{ gam}}\)
\({m_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)
\( \to C{\% _{HN{O_3}}} = \frac{{18,9}}{{558,2}} = 3,386\% \)
\(C{\% _{HCl}} = \frac{{7,3}}{{558,2}} = 1,3\% \)
