Đáp án:
\(C{\% _{HCl}} = 7,3\% \)
\( C{\% _{MgC{l_2}}} = 9,13\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
Ta có:
\({n_{MgO}} = \frac{8}{{24 + 16}} = 0,2{\text{ mol = }}{{\text{n}}_{MgC{l_2}}}\)
\( \to {m_{MgC{l_2}}} = 0,2.(24 + 35,5.2) = 19{\text{ gam}}\)
\({n_{HCl}} = 2{n_{Mg}} = 0,2.2 = 0,4{\text{ mol}}\)
\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{{m_{HCl}}}}{{{m_{dd\;{\text{HCl}}}}}} = \frac{{14,6}}{{200}}.100\% = 7,3\% \)
BTKL:
\({m_{dd\;{\text{sau phản ứng}}}} = {m_{MgO}} + {m_{dd\;{\text{HCl}}}} = 8 + 200 = 208{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{{m_{MgC{l_2}}}}}{{{m_{dd}}}}.100\% = \frac{{19}}{{208}}.100\% = 9,13\% \)
