Đáp án:
\( C{\% _{MgC{l_2}}} = 6,1\% ;C{\% _{HCl}} = 14,167\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
\({n_{MgO}} = \frac{6}{{24 + 16}} = 0,15{\text{ mol}}\)
\({m_{HCl}} = 150.19,6\% = 29,4{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{29,4}}{{36,5}} = 0,805{\text{ mol > 2}}{{\text{n}}_{MgO}}\)
Vậy \(HCl\) dư
\( \to {n_{MgC{l_2}}} = {n_{MgO}} = 0,1{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,1.(24 + 35,5.2) = 9,5{\text{ gam}}\)
\({m_{HCl{\text{ dư}}}} = 29,4 - 0,1.2.36,5 = 22,1{\text{ gam}}\)
\({m_{dd}} = {m_{MgO}} + {m_{dd\;{\text{HCl}}}} = 150 + 6 = 156{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{9,5}}{{156}} = 6,1\% ;C{\% _{HCl}} = \frac{{22,1}}{{156}} = 14,167\% \)
