Đáp án:
\(\% {m_{Mg}} = 54,5\% ; \% {m_{CuO}} = 45,5\% \)
\({V_{HCl}} = V_{NaOH}= 0,25{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{Mg}}\)
\( \to {m_{Mg}} = 0,2.24 = 4,8{\text{ gam}}\)
\( \to {m_{CuO}} = 8,8 - 4,8 = 4{\text{ gam}}\)
\( \to {n_{CuO}} = \frac{4}{{64 + 16}} = 0,05{\text{ mol}}\)
\( \to \% {m_{Mg}} = \frac{{4,8}}{{8,8}} = 54,5\% \to \% {m_{CuO}} = 45,5\% \)
\({n_{HCl}} = 2{n_{Mg}} + 2{n_{CuO}} = 0,2.2 + 0,05.2 = 0,5{\text{ mol}}\)
\( \to {V_{HCl}} = \frac{{0,5}}{2} = 0,25{\text{ lít}}\)
Cho \(B\) tác dụng với \(NaOH\)
\(MgC{l_2} + 2NaOH\xrightarrow{{}}Mg{(OH)_2} + 2NaCl\)
\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)
\( \to {n_{NaOH}} = 2{n_{NaCl}} = {n_{HCl}} = 0,5{\text{ mol}}\)
\( \to {V_{NaOH}} = \frac{{0,5}}{2} = 0,25{\text{ lít}}\)
