Đáp án:
73,5%
Giải thích các bước giải:
\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl(1)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O(2)\\
nBaC{l_2} = \dfrac{{104}}{{208}} = 0,5\,mol\\
n{H_2}S{O_4}(1) = nBaC{l_2} = 0,5\,mol\\
m{\rm{dd}}NaOH = 250 \times 1,28 = 320g\\
nNaOH = \dfrac{{320 \times 25\% }}{{40}} = 2\,mol\\
n{H_2}S{O_4}(2) = \dfrac{2}{2} = 1\,mol\\
n{H_2}S{O_4} = 1 + 0,5 = 1,5\,mol\\
C\% {H_2}S{O_4} = \dfrac{{1,5 \times 98}}{{200}} \times 100\% = 73,5\%
\end{array}\)