Đáp án:
\({m_{AgCl}}= 143,5{\text{ gam}}\)
\({C_{M{\text{ NaN}}{{\text{O}}_3}}} = 0,2M;{C_{M{\text{ AgN}}{{\text{O}}_3}}} = 0,4M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaCl + AgN{O_3}\xrightarrow{{}}AgCl + NaN{O_3}\)
Ta có:
\({n_{NaCl}} = \frac{{5,85}}{{23 + 35,5}} = 0,1{\text{ mol}}\)
\({n_{AgN{O_3}}} = \frac{{34}}{{170}} = 0,2{\text{ mol > }}{{\text{n}}_{NaCl}}\)
Vậy \(AgNO_3\) dư
\( \to {n_{AgCl}} = {n_{NaCl}} = 0,1{\text{ mol}}\)
\( \to {m_{AgCl}} = 0,1.(108 + 35,5) = 143,5{\text{ gam}}\)
Ta có:
\({V_{dd}} = 300 + 200 = 500{\text{ ml = 0}}{\text{,5 lít}}\)
\({n_{NaN{O_3}}} = {n_{NaCl}} = 0,1{\text{ mol;}}{{\text{n}}_{AgN{O_3}{\text{ dư}}}} = 0,3 - 0,1 = 0,2{\text{ mol}}\)
\( \to {C_{M{\text{ NaN}}{{\text{O}}_3}}} = \frac{{0,1}}{{0,5}} = 0,2M;{C_{M{\text{ AgN}}{{\text{O}}_3}}} = \frac{{0,2}}{{0,5}} = 0,4M\)