Câu 1:

$m_{Ba(OH)_2}= 40.\dfrac{34,2}{100}=13,68g$

$n_{Ba(OH)_2}=\dfrac{13,68}{171}=0,08mol$

$Ba(OH)_2 + 2HCl ---> BaCl_2 + 2H_2O$

     0,08             0,16                   0,08                          mol

$m_{HCl}= 0,16.36,5=5,84g$

$m_{ddHCl}=5,84:$ $\dfrac{7,3}{100}=80g$

Bảo toàn khối lượng:

$m_{ddBa(OH)_2}+ m_{ddHCl}= m_{dd sau}$

<=> $m_{dd sau}=40+80=120g$

Câu 2: 

$n_{Ba(OH)_2}=\dfrac{17,1}{171}=0,1mol$

$Ba(OH)_2 + H_2SO_4 ---> BaSO_4↓ + 2H_2O$

     0,1                  0,1                      0,1                          mol

Bảo toàn khối lượng:

$m_{ddBa(OH)_2}+m_{ddH_2SO_4}-m_{BaSO_4}= m_{dd sau}$

<=> $m_{dd sau}=17,1+200-0,1.233=193,8g$

C% $_{ddH_2SO_4}=\dfrac{0,1.98}{200}=4,9$ %

Câu 3: 

$n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1mol$

$ Na_2CO_3 + 2HCl ---> 2NaCl + H_2O +CO_2$

         0,1            0,2                    0,2                           0,1     mol

$m_{HCl}= 0,2.36,5=7,3g$

C%$_{ddHCl}=\dfrac{7,3}{200}=$$3,65$%

Bảo toàn khối lượng:

$m_{Na_2CO_3}+m_{ddHCl}-m{CO_2}= m_{dd sau}$

<=> $m_{dd sau}=10,6+200-0,1.44=206,2g$

C%$_{ddNaCl}=\dfrac{0,2.58,5}{206.2}=$$5,67$%

$1. \\m_{Ba(OH)_2}=40.34,2\%=13,68g \\⇒n_{Ba(OH)_2}=\dfrac{13,68}{171}=0,08mol \\PTHH : \\Ba(OH)_2+2HCl\to BaCl_2+H_2O \\a.Theo\ pt : \\n_{HCl}=2.n_{Ba(OH)_2}=2.0,08=0,16mol \\⇒m_{HCl}=0,16.36,5=5,84g \\m_{dd\ HCl}=\dfrac{5,84}{7,3\%}=80g \\b.m_{dd\ spu}=40+80=120g \\c.Theo\ pt : \\n_{BaCl_2}=n_{Ba(OH)_2}=0,08mol \\⇒m_{BaCl_2}=0,08.208=16,64g \\⇒C\%_{BaCl_2}=\dfrac{16,64}{120}.100\%=13,86\% \\3. \\n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1mol \\PTHH : \\Na_2CO_3+2HCl\to 2NaCl+2H_2O+CO_2↑ \\a.Theo\ pt : \\n_{HCl}=2.n_{Na_2CO_3}2.0,1=0,2mol \\⇒m_{HCl}=0,2.36,5=7,3g \\⇒C\%_{HCl}=\dfrac{7,3}{200}.100\%=3,65\% \\b.Theo\ pt : \\n_{NaCl}=2.n_{Na_2CO_3}=2.0,1=0,2mol \\⇒m_{NaCl}=0,2.58,5=11,7g \\n_{CO_2}=n_{Na_2CO_3}=0,1mol \\m_{dd\ spu}=10,6+200-0,1.44=206,2g \\⇒C\%_{NaCl}=\dfrac{11,7}{206,2}.100\%=5,67\%$