Giải thích các bước giải:

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2HCl\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 100 \times 1,14 = 114g\\
 \to {m_{{H_2}S{O_4}}} = \dfrac{{114 \times 20\% }}{{100\% }} = 22,8g\\
 \to {n_{{H_2}S{O_4}}} = 0,23mol\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2\% }}{{100\% }} = 20,8g\\
 \to {n_{BaC{l_2}}} = 0,1mol\\
 \to {n_{BaC{l_2}}} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{{H_2}S{O_4}(pt)}} = {n_{BaC{l_2}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}(dư)}} = 0,13mol\\
a)\\
{n_{{\rm{BaS}}{O_4}}} = {n_{BaC{l_2}}} = 0,1mol\\
 \to {m_{{\rm{BaS}}{O_4}}} = 23,3g\\
b)\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol \to {m_{HCl}} = 7,3g\\
{m_{{H_2}S{O_4}(dư)}} = 12,74g\\
{m_{{\rm{dd}}thuđược}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{{\rm{ddBaC}}{{\rm{l}}_2}}} - {m_{{\rm{BaS}}{O_4}}} = 490,7g\\
 \to C{\% _{HCl}} = \dfrac{{7,3}}{{490,7}} \times 100\%  = 1,5\% \\
 \to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{12,74}}{{490,7}} \times 100\%  = 2,6\% 
\end{array}\)

a,

$m_{dd H_2SO_4}=100.1,14=114g$

$\Rightarrow n_{H_2SO_4}=\dfrac{114.20\%}{98}=0,23(mol)$

$n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1(mol)$

$BaCl_2+H_2SO_4\to BaSO_4+2HCl$

$\Rightarrow n_{BaSO_4}=0,1(mol)$

$m_{\downarrow}=0,1.233=23,3g$

b,

$m_{dd\text{spu}}=114+400-23,3=490,7g$

$n_{HCl}=0,1.2=0,2(mol)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5.100}{490,7}=1,49\%$

$n_{H_2SO_4\text{dư}}=0,23-0,1=0,13(mol)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,13.98.100}{490,7}=2,6\%$