Đáp án:
\(m = 14,35{\text{ gam}}\)
\({C_{M{\text{ NaN}}{{\text{O}}_3}}} = {C_{M{\text{ AgN}}{{\text{O}}_3}{\text{ dư}}}} = 0,2M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaCl + AgN{O_3}\xrightarrow{{}}AgCl + NaN{{\text{O}}_3}\)
Ta có:
\({n_{NaCl}} = \dfrac{{5,85}}{{23 + 35,5}} = 0,1{\text{ mol}}\)
\({n_{AgN{O_3}}} = \dfrac{{34}}{{108 + 62}} = 0,2{\text{ mol > }}{{\text{n}}_{NaCl}}\)
Vậy \(AgNO_3\) dư.
\( \to {n_{AgCl}} = {n_{NaCl}} = {n_{NaN{O_3}}} = 0,1{\text{ mol}}\)
\( \to m = {m_{AgCl}} = 0,1.(108 + 35,5) = 14,35{\text{ gam}}\)
\({n_{AgN{O_3}{\text{ dư}}}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)
\( \to {V_{dd}} = 300 + 200 = 500{\text{ ml = 0}}{\text{,5 lít}}\)
\( \to {C_{M{\text{ NaN}}{{\text{O}}_3}}} = {C_{M{\text{ AgN}}{{\text{O}}_3}{\text{ dư}}}} = \dfrac{{0,1}}{{0,5}} = 0,2M\)