Đáp án:
\(V=83,33 ml\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgC{O_3} + 2HCl\xrightarrow{{}}MgC{l_2} + C{O_2} + {H_2}O\)
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
Ta có:
\({n_{MgC{O_3}}} = \frac{{4,2}}{{24 + 60}} = 0,05{\text{ mol}}\)
\({n_{A{l_2}{O_3}}} = \frac{{5,1}}{{27.2 + 16.3}} = 0,05{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{MgC{O_3}}} + 6{n_{A{l_2}{O_3}}} = 0,05.2 + 0,05.6 = 0,4{\text{ mol}}\)
\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{14,6}}{{14,6\% }} = 100{\text{ gam}}\)
\( \to V = {V_{{\text{dd HCl}}}} = \frac{{100}}{{1,2}} = 83,33{\text{ ml}}\)
