1)
Phản ứng xảy ra:
\(2Al + 6{H_2}S{O_4}\xrightarrow{{{t^o}}}A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)
\(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)
Ta có:
\({n_{S{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)
\( \to {n_{Al}} = \frac{2}{3}{n_{S{O_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{Al}} = 0,1.27 = 2,7{\text{ gam}} \to {{\text{m}}_{A{l_2}{O_3}}} = 10,2{\text{ gam}}\)
\( \to {n_{A{l_2}{O_3}}} = \frac{{10,2}}{{27.2 + 16.3}} = 0,1{\text{ mol}}\)
\( \to {n_{{H_2}S{O_4}}} = 3{n_{Al}} + \frac{3}{2}{n_{A{l_2}{O_3}}} = 0,1.3 + \frac{3}{2}.0,1 = 0,45{\text{ mol}}\)
\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} + {n_{A{l_2}{O_3}}} = \frac{1}{2}.0,1 + 0,1 = 0,15{\text{ mol}}\)
\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,15.(27.2 + 96.3) = 51,3{\text{ gam}}\)
2)
Phản ứng xảy ra:
\(S{O_2} + B{r_2} + 2{H_2}O\xrightarrow{{}}{H_2}S{O_4} + 2HBr\)
\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2HCl\)
Ta có:
\({n_{BaS{O_4}}} = \frac{{46,6}}{{137 + 32 + 16.4}} = 0,2{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}} = {n_{S{O_2}}}\)
\( \to V = {V_{S{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
