Đáp án:
\( \to {m_{Mg{{(OH)}_2}}}= 5,8{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgC{l_2} + 2KOH\xrightarrow{{}}Mg{(OH)_2} + 2KCl\)
Ta có:
\({n_{KOH}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}}\)
\( \to {n_{Mg{{(OH)}_2}}} = \frac{1}{2}{n_{KOH}} = 0,1{\text{ mol}}\)
\( \to {m_{Mg{{(OH)}_2}}} = 0,1.(24 + 17.2) = 5,8{\text{ gam}}\)
BTKL:
\({m_{KOH}} + {m_{dd{\text{MgC}}{{\text{l}}_2}}} = {n_{dd{\text{B}}}} + {m_{Mg{{(OH)}_2}}}\)
\( \to 11,2 + 120 = {m_B} + 5,8 \to {m_B} = 125,4{\text{ gam}}\)
\({n_{KCl}} = {n_{KOH}} = 0,2{\text{ mol}}\)
\( \to {m_{KCl}} = 0,2.(39 + 35,5) = 14,9{\text{ gam}}\)
\( \to C{\% _{KCl}} = \frac{{14,9}}{{125,4}} = 11,88\% \)
