$\begin{align}  & {{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,3.1,5=0,45\left( mol \right) \\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,3.2=0,6\left( mol \right) \\\end{align}$

  $KOH+HCl\to KCl+{{H}_{2}}O$

    0,45$\leftarrow $0,45                                         (mol)

$2KOH+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O$

     1,2 $\leftarrow $ 0,6                 (mol)

Theo phương trình ta có:

 $\begin{align}  & {{n}_{KOH}}={{n}_{HCl}}+2.{{n}_{{{H}_{2}}S{{O}_{4}}}}=0,45+2.0,6=1,65\left( mol \right) \\ & \Rightarrow {{m}_{KOH}}={{n}_{KOH}}.{{M}_{KOH}}=1,65.56=92,4\left( g \right) \\\end{align}$

$\Rightarrow $Chọn đáp án D.