Đáp án:

a)   Viết ptpứ xảy ra: `H_2SO_4+BaCl_2toBaSO_4↓+2Hcl`

   `mH_2SO_4={20\times29,4}/100=5,88(g)` 

   Suy ra `nH_2SO_4=0,06(mol)`

  `mBaCl_2={5,2\times100}/100=5,2(g)`

   Suy ra 'nBaCl_2=0,025(mol)`

  $\dfrac{n_{H_{2}SO_4} pứ}{n_{H_{2}SO_4} bđ}=$`{0,06}/1>`$\dfrac{n_{BaCl_2} pứ}{n_{BaCl_2}bđ}=$`{0,025}/1`

   Suy ra `H_2SO_4` dư, `BaCl_2` hết

   Số `nBaCl_2=nBaSO_4(=0,025mol)`

   Suy ra `mBaSO_4=0,025\times233=5,825(g)`

b)   Những chất còn lại sau pứ là `H_2SO_4` dư, `Hcl`.

   `mdd` sau pứ `=29,4+100-5,825=123,575(g)`

   Số `nH_2SO_4` dư `=0,06-0,025=0,035(mol)`

   `mH_2SO_4=0,035\times98=3,43(g)`

   Suy ra `C%H_2SO_4={mct}/{mdd}\times100={3,43}/{123,575}\times100≈2,78%`

   Số `nHcl=0,025\times2=0,05(mol)`

   `mHcl=0,05\times36,5=1,825`

   Suy ra `C%Hcl={mct}/{mdd}\times100={1,825}/{123,575}\times100≈1,47%`

$a.m_{H_2SO_4}=29,4.20\%=5,88g \\⇒n_{H_2SO_4}=\dfrac{5,88}{98}=0,06mol \\m_{BaCl_2}=5,2\%.100=5,2g \\⇒n_{BaCl_2}=\dfrac{5,2}{208}=0,025mol \\PTHH :$

                 $ H_2SO_4 + BaCl_2\to BaSO_4↓+2HCl$

Theo pt :      1 mol        1 mol

Theo đbài : 0,06 mol   0,025 mol

Tỉ lệ : $\dfrac{0,06}{1}>\dfrac{0,025}{1}$

⇒Sau phản ứng H2SO4 dư

$Theo\ pt : \\n_{BaSO_4}=n_{BaCl_2}=0,025mol \\⇒m_{BaSO_4}=0,025.233=5,825g \\b.Theo\ pt : \\n_{H_2SO_4\ pư}=n_{BaCl_2}=0,025mol \\⇒n_{H_2SO_4\ dư}=0,06-0,025=0,035mol \\⇒m_{H_2SO_4\ dư}=0,035.98=3,43g \\n_{HCl}=2.n_{BaCl_2}=2.0,025=0,05mol \\⇒m_{HCl}=0,05.36,5=1,825g \\m_{dd\ spu}=29,4+100-5,825=123,575g \\⇒C\%_{H_2SO_4\ dư}=\dfrac{3,43}{123,575}.100\%=2,78\% \\C\%_{HCl}=\dfrac{1,825}{123,575}.100\%=1,48\%$