a,

$m_{dd BaCl_2}=400.1,03=412g$

$\Rightarrow n_{BaCl_2}=\dfrac{412.5,2\%}{208}=0,103 mol$

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

$\Rightarrow n_{H_2SO_4}=\dfrac{114.20\%}{98}=0,233 mol$

$BaCl_2+H_2SO_4\to BaSO_4+2HCl$

$\Rightarrow n_{BaSO_4}=0,103 mol$

$m_{BaSO_4}=0,103.233=24g$

b,

$m_{dd spứ}=412+114-24=502g$

$n_{HCl}=0,103.2=0,206 mol$

$\Rightarrow C\%_{HCl}=\dfrac{0,206.36,5.100}{502}=1,5\%$

$n_{H_2SO_4\text{dư}}=0,233-0,103=0,13 mol$

$\Rightarrow C\%_{H_2SO_4}=\dfrac{0,13.98.100}{502}=2,54\%$

a.

-mdd BaCl2=400.1,03=412 g

m ct BaCl2=$\frac{412.5,2}{100}$ =21,424 g

n BaCl2=$\frac{21,424}{208}$ =0,103 mol

-mdd H2SO4=100.1,14=114 g

mdd H2SO4=$\frac{144.20}{100}$ =22,8 g

n H2SO4=$\frac{22,8}{98}$ ≈0,23 mol

           BaCl2+H2SO4→BaSO4↓+2HCl

trc pứ:0,103     0,23                                 mol 

pứ:      0,103    0,103     0,103      0,206   mol

sau pứ:  0        0,127     0,103       0,206   mol

m kết tủa=m BaSO4=0,103.233≈24 g

b.

-dd sau phản ứng gồm:H2SO4 dư,HCl

-mdd sau=m BaCl2+m H2SO4-m BaSO4

              = 412+114 -24 =502 g

m ct HCl=0,206.36,5=7,519 g

m ct H2SO4 dư=0,127.98=12,446 g

C% HCl=$\frac{7,519}{502}$ .100≈1,5%

C% H2SO4 dư=$\frac{12,446}{502}$ .100≈2,48 %

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