Đáp án:

\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 133,3g\\
b)\\
C{\% _{MgS{O_4}}} = 16,985\% \\
c)\\
{V_{NaOH}} = 267ml
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{MgO}} = \dfrac{m}{M} = \dfrac{8}{{40}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {n_{MgO}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,2 \times 98 = 19,6g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{19,6 \times 100}}{{14,7}} = 133,3g\\
b)\\
{n_{MgS{O_4}}} = {n_{MgO}} = 0,2mol\\
{m_{MgS{O_4}}} = n \times M = 0,2 \times 120 = 24g\\
{m_{{\rm{dd}}spu}} = 8 + 133,3 = 141,3g\\
C{\% _{MgS{O_4}}} = \dfrac{{24}}{{141,3}} \times 100\%  = 16,985\% \\
c)\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,4mol\\
{V_{NaOH}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,4}}{{1,5}} = 0,267l = 267ml
\end{array}\)