4)
Phản ứng xảy ra:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
\({n_{C{l_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{Mn{O_2}}}\)
\( \to {m_{Mn{O_2}}} = 0,3.(55 + 16.2) = 26,1{\text{ gam}}\)
5)
Phản ứng xảy ra:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
Ta có:
\({n_{Mn{O_2}}} = \frac{{17,4}}{{55 + 16.2}} = 0,2{\text{ mol = }}{{\text{n}}_{C{l_2}}}\)
\( \to {V_{C{l_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)