Đáp án:

 a) cần 90 gam dung dịch NaOH

b) cần 861,24 ml dung dịch KOH

Giải thích các bước giải:

\[\begin{gathered}
  a)2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O \hfill \\
  {n_{{H_2}S{O_4}}} = 0,3.1,5 = 0,45{\text{ mol}} \to {{\text{n}}_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,9{\text{ mol}} \hfill \\
   \to {{\text{m}}_{NaOH}} = 0,9.40 = 36{\text{ gam}} \hfill \\
   \to {{\text{m}}_{dd{\text{NaOH}}}} = \frac{{36}}{{40\% }} = 90{\text{ gam}} \hfill \\
  b)2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O \hfill \\
   \to {n_{KOH}} = 0,9{\text{ mol}} \to {{\text{m}}_{KOH}} = 0,9.56 = 50,4{\text{ gam}} \hfill \\
   \to {{\text{m}}_{{\text{dd KOH}}}} = \frac{{50,4}}{{5,6\% }} = 900{\text{gam}} \hfill \\
   \to {{\text{V}}_{dd{\text{ KOH}}}} = \frac{{900}}{{1,045}} = 861,24{\text{ ml}} \hfill \\ 
\end{gathered} \]

a,

nH2SO4= 0,3.1,5= 0,45 mol  

H2SO4+ 2NaOH -> Na2SO4+ 2H2O 

=> nNaOH= 0,9 mol 

mNaOH= 0,9.40= 36g 

=> mdd NaOH= 36 : 40%= 90g 

b, 

H2SO4+ 2KOH -> K2SO4+ 2H2O 

=> nKOH= 0,9 mol 

mKOH= 56.0,9= 50,4g 

=> mdd KOH= 50,4 : 5,6%= 900g 

=> V dd KOH= $\frac{900}{1,045}$= 861,24 ml