Bài 9:

m ct NaOH=$\frac{10.100}{100}$=10 g

⇒n NaOH=$\frac{10}{40}$=0,25 mol

2NaOH+CuCl2→2NaCl+Cu(OH)2↓

 0,25  →               0,25      0,125            mol

m kết tủa=m Cu(OH)2=0,125.98=12,25 g

mdd sau=mdd NaOH+mdd CuCl2-mCu(OH)2

              =100 + 33,75 - 12,25 =121,5 g

m ct NaCl=0,25.58,5=14,625 g

⇒C% NaCl=$\frac{14,625}{121,5}$.100≈12,037 %

Bài 10:

n Na2SO4=0,1.0,5=0,05 mol

Na2SO4+BaCl2→BaSO4↓+2NaCl

 0,05→    0,05         0,05       0,1            mol

a.

m kết tủa=m BaSO4=0,05.233=11,65 g

b.

CM BaCl2=$\frac{0,05}{0,15}$=$\frac{1}{3}$ M

c.

-Vì thể tích dung dịch thay đổi không đáng kể:

⇒Vdd sau pứ=Vdd ban đầu=0,1+0,15=0,25 l

CM NaCl=$\frac{0,1}{0,25}$=0,4 M

---------------------Nguyễn Hoạt------------------------

9)

Phản ứng xảy ra:

\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)

Ta có: 

\({m_{NaOH}} = 100.10\%  = 10{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{10}}{{40}} = 0,25{\text{ mol}}\)

\({n_{CuC{l_2}}} = \frac{{33,75}}{{64 + 35,5.2}} = 0,25{\text{ mol > }}\frac{1}{2}{n_{NaOH}} \to {n_{CuC{l_2}}}\) dư.

\( \to {n_{Cu{{(OH)}_2}}} = \frac{1}{2}{n_{NaOH}} = 0,125{\text{ mol}} \to {m_{Cu{{(OH)}_2}}} = 0,125.(64 + 17.2) = 12,25{\text{ gam}}\)

Dung dịch sau phản ứng chứa \({n_{NaCl}} = {n_{NaOH}} = 0,25{\text{ mol; }}{{\text{n}}_{CuC{l_2}{\text{ dư}}}} = 0,25 - 0,125 = 0,125{\text{ mol}}\)

BTKL: \({m_{dd{\text{ sau phản ứng}}}} = {m_{dd{\text{NaOH}}}} + {m_{CuC{l_2}}} - {m_{Cu{{(OH)}_2}}} = 100 + 33,75 - 12,25 = 121,5{\text{ gam}}\)

\({m_{NaCl}} = 0,25.58,5 = 14,625{\text{ gam; }}{{\text{m}}_{CuC{l_2}}} = 0,125.135 = 16,875{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaCl}} = \frac{{14,625}}{{121,5}} = 12,04\% ;C{\% _{CuC{l_2}}} = \frac{{16,875}}{{121,5}} = 13,89\% \)

\({\text{C}}{{\text{\% }}_{NaCl}} = 12,04\% ;C{\% _{CuC{l_2}}} = 13,89\% \)

10)

Phản ứng xảy ra:

\(N{a_2}S{O_4} + BaC{l_2}\xrightarrow{{}}2NaCl + BaS{O_4}\)

Ta có: 

\({n_{N{a_2}S{O_4}}} = 0,1.0,5 = 0,05{\text{ mol = }}{{\text{n}}_{BaS{O_4}}} = {n_{BaC{l_2}}} \to {m_{BaS{O_4}}} = 0,05.233 = 11,65{\text{ gam}}\)

\({C_{M{\text{ BaC}}{{\text{l}}_2}}} = \frac{{0,05}}{{0,15}} = 0,333M\)

Sau phản ứng dung dịch chứa 

\({n_{NaCl}} = 2{n_{N{a_2}S{O_4}}} = 0,1{\text{ mol;}}{{\text{V}}_{dd}} = 100 + 150 = 250ml = 0,25{\text{ lít}} \to {{\text{C}}_{{\text{M }}NaCl}} = \frac{{0,1}}{{0,25}} = 0,4M\)