$MgO+{{H}_{2}}S{{O}_{4}}\to MgS{{O}_{4}}+{{H}_{2}}O$

0,75 $\leftarrow $ 0,75                       (mol)

${{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,3.2,5=0,75\left( mol \right)$

Theo phương trình ta có:

$\begin{align} & {{n}_{MgO}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,75\left( mol \right) \\ & \Rightarrow {{m}_{MgO}}={{n}_{MgO}}.{{M}_{MgO}}=0,75.40=30\left( g \right) \\\end{align}$

$\Rightarrow $Chọn đáp án B