1)

\({n_{NaOH}} = 0,15.0,5 = 0,075{\text{ mol}}\)

2)

\({m_{N{a_2}C{O_3}}} = 120.5,3\%  = 6,36{\text{ gam}}\)

\( \to {n_{N{a_2}C{O_3}}} = \frac{{6,36}}{{23.2 + 60}} = 0,06{\text{ mol}}\)

3)

a)

\(C{\% _{CuS{O_4}}} = \frac{{20}}{{200}} = 10\% \)

b)

\(C{\% _{NaCl}} = \frac{{12}}{{180}} = 6,67\% \)

c)

\({n_{KN{O_3}}} = \frac{{10,1}}{{39 + 14 + 16.3}} = 0,1{\text{ mol}}\)

\( \to {C_{M{\text{ KN}}{{\text{O}}_3}}} = \frac{{0,1}}{{0,3}} = 0,333M\)

d)

\({n_{KCl}} = \frac{{14,9}}{{39 + 35,5}} = 0,2{\text{ mol}}\)

\( \to {C_{M{\text{ KCl}}}} = \frac{{0,2}}{{0,5}} = 0,4M\)

1)  n N a O H = 0 , 15.0 , 5 = 0 , 075  mol

2)  m N a 2 C O 3 = 120.5 , 3 % = 6 , 36  gam → n N a 2 C O 3 = 6 , 36 23.2 + 60 = 0 , 06  mol

3)  a)  C % C u S O 4 = 20 200 = 10 %  

b)  C % N a C l = 12 180 = 6 , 67 %

c)  n K N O 3 = 10 , 1 39 + 14 + 16.3 = 0 , 1  mol → C M  KN O 3 = 0 , 1 0 , 3 = 0 , 333 M

d)  n K C l = 14 , 9 39 + 35 , 5 = 0 , 2  mol → C M  KCl = 0 , 2 0 , 5 = 0 , 4 M