Đáp án:
\(C{\% _{MgC{l_2}}} = 11,99\% \)
Giải thích các bước giải:
Ta có:
\(C{\% _{MgC{l_2}}} = \frac{{{m_{MgC{l_2}}}}}{{{m_{dd}}}} = \frac{{23,75}}{{198}} = 11,99\% \)
Cho dung dịch \(AgNO_3\) vào dung dịch \(A\)
\(MgC{l_2} + 2AgN{O_3}\xrightarrow{{}}2AgCl + Mg{(N{O_3})_2}\)
Ta có:
\({n_{MgC{l_2}}} = \frac{{23,75}}{{24 + 35,5.2}} = 0,25{\text{ mol}}\)
\({m_{AgN{O_3}}} = 170.10\% = 17{\text{ gam}}\)
\( \to {n_{AgN{O_3}}} = \frac{{17}}{{108 + 62}} = 0,1{\text{ mol < 2}}{{\text{n}}_{MgC{l_2}}}\)
\( \to {n_{AgCl}} = {n_{AgN{O_3}}} = 0,1{\text{ mol}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = 198 + 170 - {m_{AgCl}}\)
\( = 198 + 170 - 0,1.(108 + 35,5) = 353,65{\text{ gam}}\)
\({n_{MgC{l_2}{\text{ }}dư}} = 0,25 - \frac{{0,1}}{2} = 0,2{\text{ mol}}\)
\({n_{Mg{{(N{O_3})}_2}}} = \frac{1}{2}{n_{AgN{O_3}}} = 0,05{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,2.(24 + 35,5.2) = 19{\text{ gam}}\)
\({m_{Mg{{(N{O_3})}_2}}} = 0,05.(24 + 62.2) = 7,4{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{19}}{{353,65}} = 5,37\% \)
\(C{\% _{Mg{{(N{O_3})}_2}}} = \frac{{7,4}}{{353,65}} = 2,09\% \)
