$\text{a.}$
$2SO_{2}$ + $O_{2}$ $\xrightarrow[V_{2}O_{5}]{t^{0}}$ $2SO_{3}$
$SO_{3}$ + $H_{2}O$ → $H_{2}SO_{4}$
$\text{$n_{SO_{3}}$ = $n_{H_{2}SO_{4}}$ = 300/1000*1 = 0.3 mol}$
$\text{$n_{SO_{2}}$ = $n_{SO_{3}}$ = 0.3 mol ⇒ $m_{SO_{2}}$ = a = 64*0.3 = 19.2 g}$
$\text{b.}$
$\text{$n_{Al_{2}O_{3}}$ = 15.3/102 = 0.15 mol}$
$Al_{2}O_{3}$ + $3H_{2}SO_{4}$ → $Al_{2}(SO_{4})_{3}$ + $3H_{2}O$
$\text{Ta có tỉ lệ: $\frac{0.15}{1}$ > $\frac{0.3}{3}$ ⇒ $n_{Al_{2}O_{3}}$ dư}$
$\text{$n_{Al_{2}(SO_{4})_{3}}$ = 1/3$n_{H_{2}SO_{4}}$ = 1/3*0.3 = 0.1 mol}$
$\text{⇒ $C_{M}$ = 0.1/0.3 = 1/3 M}$