Đáp án:

\(\begin{array}{l}
a)\\
{m_{BaS{O_4}}} = 5,825g\\
b)\\
C{\% _{HCl}} = 1,48\% \\
C{\% _{{H_2}S{O_4}}} = 2,78\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl\\
{m_{{H_2}S{O_4}}} = \dfrac{{29,4 \times 20}}{{100}} = 5,88g\\
{n_{{H_2}S{O_4}}} = \dfrac{{5,88}}{{98}} = 0,06mol\\
{m_{BaC{l_2}}} = \dfrac{{100 \times 5,2}}{{100}} = 5,2g\\
{n_{BaC{l_2}}} = \dfrac{{5,2}}{{208}} = 0,025mol\\
\dfrac{{0,06}}{1} > \dfrac{{0,025}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,025mol\\
{m_{BaS{O_4}}} = 0,025 \times 233 = 5,825g\\
b)\\
{m_{ddspu}} = 29,4 + 100 - 5,825 = 123,575g\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,05mol\\
{m_{HCl}} = 0,05 \times 36,5 = 1,825g\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - {n_{BaC{l_2}}} = 0,035mol\\
{m_{{H_2}S{O_4}d}} = 0,035 \times 98 = 3,43g\\
C{\% _{HCl}} = \dfrac{{1,825}}{{123,575}} \times 100\%  = 1,48\% \\
C{\% _{{H_2}S{O_4}}} = \dfrac{{3,43}}{{123,575}} \times 100\%  = 2,78\% 
\end{array}\)

`mH_2SO_4= 29,4.20%=5,88`(g)

  `=> nH_2SO_4=0,06` (mol)

`mBaCl_2=100.5,2%=5,2` (g)

  `=> nBaCl_2 =0,025` (mol)

a, `BaCl_2 + H_2SO_4 -> BaSO_4↓ + 2HCl`

 `nBaCl_2 < nH_2SO_4 (0,025<0,06 )`

 `=> BaCl_2` hết `; H_2SO_4` dư

 `=> nBaSO_4 = nBaCl_2 = 0,025` (mol)

 `=> mBaSO_4=5,825` (g)

b, 

 `mdd = m`dd `BaCl_2 + m`dd `H2SO_4 - mBaSO_4`

          `= 29,4+100-5,825 `

          `=123,575`

 `nHCl = 2.nBaCl_2= 0,05` (mol)

 `=> mHCl = 1,825` (g)

`nH_2SO_4` pứ `= nBaCl2 = 0,025` (mol)

`mH_2SO_4` dư `= (0,06-0,025).98=3,43` (g)

`% H_2SO_4=(3,43)/(123,575).100%=2,776%`

`%HCl = (1,825)/(123,575).100%=1,477%`