Em tham khảo nha :
\(\begin{array}{l}
a)\\
2NaOH + S{O_3} \to N{a_2}S{O_4} + {H_2}O\\
{n_{S{O_3}}} = \dfrac{{2,8}}{{22,4}} = 0,125mol\\
{n_{N{a_2}S{O_4}}} = {n_{S{O_3}}} = 0,125mol\\
{m_{N{a_2}S{O_4}}} = 0,125 \times 142 = 17,75g\\
b)\\
{n_{NaOH}} = 2{n_{S{O_3}}} = 0,25mol\\
{m_{NaOH}} = 0,25 \times 40 = 10g\\
C{\% _{NaOH}} = \dfrac{{10}}{{480}} \times 100\% = 2,0833\% \\
c)\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{17,75}}{{480 + 2,8}} \times 100\% = 3,676\%
\end{array}\)