\(n_{NaOH}=0,3.2=0,6\left(mol\right)\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,6------->0,3--------->0,3
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{29,4.100}{19,6}=150\left(g\right)\)
\(m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\)