Question

15 , 2 g  de NaOH SI koh care se gasesc in raport molar de 1 la 2 se neutralizeaza cu o solutie de acid sulfuric de concentratie 49 % . Se cere : compozitia in procente de masa a amestecului de baze , masa de solutie de acid sulfuric necesar.

Answer 1

2 NaOH + H2SO4 ---> Na2SO4 + 2 H2O
2 moli NaOH............1 mol H2SO4
a moli NaOH............b moli H2SO4
a = 2b
2 KOH + H2SO4 ---> K2SO4 + H2O
2 moli KOH..........1 mol H2SO4
c moli NaOH...........d moli H2SO4
c = 2d
a/c = 1/2 => 2a = c
Sistem
40a+56c = 15,2
2a = c
=>40a+56*2a=15,2=>152a=15,2=>a=0,1 moli
2a = c => c = 0,2 moli
masa KOH = 0.2 * 56 = 11,2 g
masa NaOH = 0,1 * 40 = 40 g
% NaOH = 4/15.2 * 100 = 26,316%
% KOH = 11,2/15,2 * 100 = 73,684%
Din ecuatia de mai sus => b = a/2 = 0,05 moli
si d = c/2 = 0,1 moli 
=> nr moli H2SO4 total = 0,15 moli = 0,15 * 98 g = 14,7 g H2SO4
c% = md/ms * 100 => ms = 14.7/49 * 100 = 300 g