Đáp án:
\({m_{AgCl}} = 28,7{\text{ gam}}\)
\(C{\% _{NaN{O_3}}} = 4,578\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(AgN{O_3} + NaCl\xrightarrow{{}}AgCl + NaN{O_3}\)
Ta có:
\({m_{AgN{O_3}}} = 200.17\% = 34{\text{ gam}}\)
\( \to {n_{AgN{O_3}}} = \frac{{34}}{{108 + 62}} = 0,2{\text{ mol = }}{{\text{n}}_{AgCl}} = {n_{NaN{O_3}}}\)
\( \to {m_{AgCl}} = 0,2.(108 + 35,5) = 28,7{\text{ gam}}\)
BTKL:
\({m_{dd{\text{ AgN}}{{\text{O}}_3}}} + {m_{dd\;{\text{NaCl}}}} = {m_{dd}} + {m_{AgCl}}\)
\( \to 200 + 200 = {m_{dd}} + 28,7 \to {m_{dd}} = 371,3{\text{ gam}}\)
\({m_{NaN{O_3}}} = 0,2.(23 + 62) = 17{\text{ gam}}\)
\( \to C{\% _{NaN{O_3}}} = \frac{{17}}{{371,3}}.100\% = 4,578\% \)