$3Ba(OH)_2+Al_2(SO_4)_3\rightarrow 3BaSO_4\downarrow+2Al(OH)_3\downarrow$ $3V_1 V_1$
$\Rightarrow [Ba(OH)_2]=[Al_2(SO_4)_3]=x (M)$
Giả sử $:\begin{cases}V_1=1 lit \\x=1 (M) \end{cases}\Rightarrow m=1(3.233+2.78) gam$
$*)V_2<3V_1\Rightarrow m_1=0,9m=0,9(3.233+2.78)\Rightarrow \frac{V_2}{V_1}=\frac{0,9.3}{1}=2,7$
$*)V_2>3V_1\Rightarrow m_2=1.3.233+78n_{Al(OH)_3}=0,9(3.233+2.78)\Rightarrow n_{Al(OH)_3}\approx 0,9$
$n_{Ba(OH)_2}=3+\frac{2-0,9}{2}=3,55\Rightarrow \frac{V_2}{V_1}=3,55$
$\Rightarrow A:$là đáp án đúng !