Đáp án:
\(\% {m_{Mg}} = 40\% ; \% {m_{MgO}} = 60\% \)
\( {V_{dd{\text{ }}{\text{HCl}}}} = 61,9196{\text{ ml}}\)
\( C{\% _{MgC{l_2}}} = 24\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Mg}}\)
\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \to {{\text{m}}_{MgO}} = 6 - 2,4 = 3,6{\text{ gam}}\)
\( \to \% {m_{Mg}} = \frac{{2,4}}{6} = 40\% \to \% {m_{MgO}} = 60\% \)
Ta có:
\({n_{MgO}} = \frac{{3,6}}{{24 + 16}} = 0,09{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,1.2 + 0,09.2 = 0,38{\text{ mol}}\)
\( \to {m_{HCl}} = 0,38.36,5 = 13,87{\text{ gam}}\)
\( \to {m_{dd\;{\text{HCl}}}} = \frac{{13,87}}{{20\% }} = 69,35{\text{ gam}}\)
\( \to {V_{dd{\text{ }}{\text{HCl}}}} = \frac{{69,35}}{{1,12}} = 61,9196{\text{ ml}}\)
BTKL:
\({m_{hh}} + {m_{dd\;{\text{HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 6 + 69,35 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 75,15{\text{ gam}}\)
\({n_{Mg{\text{C}}{{\text{l}}_2}}} = {n_{Mg}} + {n_{MgO}} = 0,1 + 0,09 = 0,19{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,19.(24 + 35,5.2) = 18,05{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{18,05}}{{75,15}} = 24\% \)
