1)

Phản ứng xảy ra:

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

\(4{\text{A}}l + 3C{l_2}\xrightarrow{{{t^o}}}2{\text{A}}lC{l_3}\)

Ta có:

\({n_{Mn{O_2}}} = \frac{{31,6}}{{55 + 16.2}} = 0,3632{\text{ kmol}}\)

\( \to {n_{C{l_2}{\text{ lt}}}} = {n_{Mn{O_2}}} = 0,3632{\text{ kmol}}\)

\( \to {n_{AlC{l_3}{\text{ lt}}}} = \frac{{2{n_{C{l_2}{\text{ lt}}}}}}{3} = 0,121{\text{ kmol}}\)

Với \(H_1=80\%;H_2=75\%\)

\( \to {n_{AlC{l_3}}} = 0,121.80\% .75\%  = 0,0726{\text{ kmol}}\)

\( \to m = {m_{AlC{l_3}}} = 0,0726.(27 + 35,5.3) = 9,6921{\text{ kg = 9692}}{\text{,1 gam}}\)

2)

Nếu thay hiệu suất là \(100\%\)

Với \(H_1=100\%;H_2=75\%\)

\( \to {n_{AlC{l_3}}} = 0,121.75\%  = 0,09075{\text{ kmol}}\)

\( \to {m_{AlC{l_3}}} = 0,09075.(27 + 35,5.3) = 12,115125{\text{ kg = 12115}}{\text{,125 gam}}\)