Đáp án:
\(\% {m_{Mg}} = 13\% ;\% {m_{MgO}} = 87\% \)
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:
\({n_{Mg}} = {n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}}\)
\( \to {m_{Mg}} = 0,05.24 = 1,2{\text{ gam}}\)
\( \to {m_{MgO}} = 9,2 - 1,2 = 8{\text{ gam}} \to {{\text{n}}_{MgO}} = \frac{8}{{24 + 16}} = 0,2{\text{ mol}}\)
\( \to \% {m_{Mg}} = \frac{{1,2}}{{9,2}} = 13\% \to \% {m_{MgO}} = 87\% \)
Ta có:
\({n_{HCl}} = 0,05.2 + 0,2.2 = 0,5\)
\( \to {m_{HCl}} = 0,5.36,5 = 18,25{\text{ gam}}\)
\( \to m = {m_{dd{\text{HCl}}}} = \frac{{18,25}}{{14,6\% }} = 125{\text{ gam}}\)
