Giải thích các bước giải:
\(\begin{array}{l}
A/\\
Ba + {H_2}S{O_4} \to BaS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,2mol\\
\to {n_{Ba}} = {n_{{H_2}}} = 0,2mol \to {m_{Ba}} = 0,2 \times 137 = 27,4g\\
\to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,2mol\\
\to {m_{{H_2}S{O_4}}} = 0,2 \times 98 = 21g\\
\to {m_{{\rm{dd}}}}_{{H_2}S{O_4}} = \dfrac{{21}}{{19,6\% }} \times 100\% = 107,14g\\
{n_{BaS{O_4}}} = {n_{{H_2}}} = 0,2mol\\
\to {m_{BaS{O_4}}} = 0,2 \times 233 = 9,32g\\
\to {m_{{\rm{dd}}}}_{BaS{O_4}} = {m_{Ba}} + {m_{{\rm{dd}}}}_{{H_2}S{O_4}} - {m_{{H_2}}} = 27,4 + 107,14 - 0,2 \times 2\\
\to {m_{{\rm{dd}}}}_{BaS{O_4}} = 134,14g\\
\to C{\% _{BaS{O_4}}} = \dfrac{{9,32}}{{134,14}} \times 100\% = 6,95\%
\end{array}\)
B,
Gọi a là số mol ban đầu của \(S{O_3}\)
Sau phản ứng thu được 100g dung dịch \({H_2}S{O_4}\) 20%
\(\begin{array}{l}
{m_{S{O_3}}} = 80ag\\
\to {m_{{H_2}S{O_4}}}(20\% ) = \dfrac{{100 \times 20}}{{100}} = 20g\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
{n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = amol\\
\to {m_{{H_2}S{O_4}}} = 98ag\\
\to {m_{{H_2}S{O_4}}}(10\% ) = 20 - 98a\\
\to {m_{{H_2}S{O_4}{\rm{dd}}}}(10\% ) = \dfrac{{100 \times (20 - 98a)}}{{10}} = 10 \times (20 - 98a)\\
\to {m_{{\rm{dd}}}}sau = 80a + 10 \times (20 - 98a) = 100\\
\to a = \dfrac{1}{9} \to {m_{S{O_3}}} = \dfrac{{80}}{9}g
\end{array}\)
