Bài 1

ZnSO4 + 2KOH -> Zn(OH)2 + K2SO4

0,2           0,4             0,2              0,2

n ZNSO4 = 32,2 / 161 = 0,2 mol

m Zn(OH)2 = 0,2 . 99 = 19,8 g

m KOH = 0,4 . 56= 22,4g

m dd KOH = ( 22,4 .100%)/10% = 224g

Chất tan trong dd spu là K2SO4

m ddspu = m ZNSO4 + m dd KOH - m Zn(OH)2 = 32,2 + 224 - 19,8 = 236,4 g

m K2SO4 = 0,2 . 174 =34,8g

C% = (34,8 . 100%)/236,4 = 14,72 %

Bài 2

2NaOH + CuCl2 -> 2NaCl + Cu(OH)2

0,4            0,2          0,4              0,2

m NaOH = (20% . 100%)/200 = 10g

n NaOH = 10 / 40 = 0,4 mol

m Cu(OH)2 = 0,4 . 98 = 39,2g

m CuCl2 = 0,2 . 135 = 27g

m dd CuCl2 = (27.100%)/10%= 270g

Chất tan trong dd spu nà NaCl

m ddspu = m ddNaOH + m ddCuCl2 - mCu(OH)2 = 200 + 270 - 39,2 = 430,8 g

m NaCl = 0,4 . 58.5 = 23,4g

C%ddspu = (23,4 . 100%) / 430,8 = 5,43 %

 

Đáp án:

 \(\begin{array}{l}
1)\\
b)\\
{m_{Zn{{(OH)}_2}}} = 19,8g\\
c)\\
{C_\% }{K_2}S{O_4} = 14,72\% \\
2)\\
b)\\
{m_{Cu{{(OH)}_2}}} = 49g\\
c)\\
{C_\% }NaCl = 7,08\% 
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
1)\\
a)\\
ZnS{O_4} + 2KOH \to Zn{(OH)_2} + {K_2}S{O_4}\\
b)\\
{n_{ZnS{O_4}}} = \dfrac{{32,2}}{{161}} = 0,2mol\\
{n_{Zn{{(OH)}_2}}} = {n_{ZnS{O_4}}} = 0,2\,mol\\
{m_{Zn{{(OH)}_2}}} = 0,2 \times 99 = 19,8g\\
c)\\
{n_{KOH}} = 2{n_{ZnS{O_4}}} = 0,4\,mol\\
{m_{{\rm{dd}}KOH}} = \dfrac{{0,4 \times 56}}{{10\% }} = 224g\\
{m_{{\rm{dd}}spu}} = 32,2 + 224 - 19,8 = 236,4g\\
{C_\% }{K_2}S{O_4} = \dfrac{{0,2 \times 174}}{{236,4}} \times 100\%  = 14,72\% \\
2)\\
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
b)\\
{n_{NaOH}} = \dfrac{{200 \times 20\% }}{{40}} = 1mol\\
{n_{Cu{{(OH)}_2}}} = \dfrac{1}{2} = 0,5\,mol\\
{m_{Cu{{(OH)}_2}}} = 0,5 \times 98 = 49g\\
c)\\
{n_{CuC{l_2}}} = \dfrac{1}{2} = 0,5\,mol\\
{m_{{\rm{ddCuC}}{{\rm{l}}_2}}} = \dfrac{{0,5 \times 135}}{{10\% }} = 675g\\
{m_{{\rm{dd}}spu}} = 200 + 675 - 49 = 826g\\
{C_\% }NaCl = \dfrac{{1 \times 58,5}}{{826}} \times 100\%  = 7,08\% 
\end{array}\)