Đáp án:

\(C\% {H_2}S{O_4} = \dfrac{{{m_{{H_2}S{O_4}}}}}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{294}}{{3620}} \times 100\%  = 8,1\% \)

Giải thích các bước giải:

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
NaOH + HCl \to NaCl + {H_2}O\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O
\end{array}\)

\(\begin{array}{l}
{m_{BaC{l_2}}} = \dfrac{{3120 \times 10}}{{100}} = 312g \to {n_{BaC{l_2}}} = 1,5mol\\
{m_{NaOH{\rm{dd}}}} = d \times V = 1,28 \times 750 = 960g \to {n_{NaOH}} = \dfrac{{960 \times 25}}{{100 \times 40}} = 6mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 3mol\\
{n_{NaOH}}({H_2}S{O_4}) = 6 - 3 = 3mol \to {n_{{H_2}S{O_4}}}dư= \dfrac{{{n_{NaOH}}({H_2}S{O_4})}}{2} = 1,5mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{{H_2}S{O_4}}}(BaC{l_2}) + {n_{{H_2}S{O_4}}}dư= 1,5 + 1,5 = 3mol\\
 \to {m_{{H_2}S{O_4}}} = 3 \times 98 = 294g\\
{m_{{\rm{dd}}}} = 3120 + 500 = 3620g\\
 \to C\% {H_2}S{O_4} = \dfrac{{{m_{{H_2}S{O_4}}}}}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{294}}{{3620}} \times 100\%  = 8,1\% 
\end{array}\)