c%=md/ms*100=>md HNO3=300*21/100=63 g HNO3
NaOH + HNO3 ---> NaNO3 + H2O
40 g NaOH.......63 g HNO3
x g NaOH..........63 g HNO3 x = 40 g NaOH = md NaOH
%apa=60%=>100 g solutie.........60 g apa =>m apa = 60 g
ms = m apa + md = 40+60=100 g
c%=md/ms*100=40/100*100=40 %
Reactia este mol la mol=> din 40 g NaOH sau 63 g HNO3 se obtin 18 g apa (folosesti masele molare)