m NaOH = c% NaOH * ms NaOH = 100 * 40/100 = 40 g NaOH
ms final = ms1+ms2 = 100+200=300 g
2 NaOH + H2SO4 ---> Na2SO4 + 2 H2O
2*40 g NaOH..........142 g Na2SO4 (masa molara)
40 g NaOH..................x g Na2SO4 x = 71 g = md final
c%finala = md final/ms final * 100 = 71/300 * 100 =>c%= 23,66 %