$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 3:}$

$\text{2NaOH + $H_{2}SO_{4}$→$Na_{2}SO_{4}$ + $H_{2}O$}$

$\text{A.}$

$\text{Ta có: $n_{NaOH}$=$\dfrac{200.4}{40}$ ÷ 100= 0,2 (mol)}$

$\text{Theo PT: $n_{H_{2}SO_{4}}$=$\dfrac{1}{2}$$n_{NaOH}$=$\dfrac{1}{2}$.0,2= 0,1 (mol)}$

$\text{⇒ $m_{H_{2}SO_{4}}$=0,1.98= 9,8 (g)}$

$\text{⇒ $m_{dd_{H_{2}SO_{4}}}$=$\dfrac{9,8}{19,6}$ × 100= 50 (g)}$

$\text{B.}$

$\text{Theo PT: $n_{Na_{2}SO_{4}}$=$\dfrac{1}{2}$$n_{NaOH}$=$\dfrac{1}{2}$.0,2= 0,1 (mol)}$

$\text{⇒ $m_{Na_{2}SO_{4}}$=0,1.142= 14,2 (g)}$

$\text{⇒ C%$Na_{2}SO_{4}$=$\dfrac{14,2}{200 + 50}$ × 100%= 5,68%}$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O\\
{n_{NaOH}} = \dfrac{{200 \times 4\% }}{{40}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{9,8}}{{19,6}} \times 100 = 50g\\
b)\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{N{a_2}S{O_4}}} = 0,1 \times 142 = 14,2g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{14,2}}{{200 + 50}} \times 100\%  = 5,68\% 
\end{array}\)