\(n_{HCl}=0,1.0,2=0,02\left(mol\right)\\ n_{H_2SO_4}=0,1.0,1=0,01\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=n_{HCl}+2.n_{H_2SO_4}=0,04\left(mol\right)\\ V_{\text{dd}NaOH}=\dfrac{0,04}{0,5}=0,08\left(l\right)\\ \Rightarrow Ch\text{ọn}.A\)