Scriem reactia si calculam cantitatea de acid consumata si masa de apa rezultata (se considera ca nitrobenzenul e separat)
78g........63g..............................18g
C6H6 + HNO3 -> C6H5-NO2 + H2O
39g.......31,5g.............................9g
consideram ca solutia initiala de amestec nitrant continea a g HNO3
20/100=a/ms => ms=5a
in solutia finala: mdf=a-31,5 si msf=5a-31,5+9=5a-22,5
5/100=(a-31,5)/(5a-22,5) => 25a-112,5=100a-3150 => 75a=3037,5 => a=40,5g HNO3
ms=md*100/c=40,5*100/20=202,5g amestec sulfonitric deci a) adevarat
pentru b)
n=m/M=40,5/63=0,64285moli
=> avem 1,2857 moli H2SO4
m=n*M=1,2857*98=126g H2SO4
mamestecinitial=202,5g
mamestec rezidual=202,5-31,5+9=180g solutie
cacidsulfuric=126*100/180=70% ≠ 63,5% deci b) fals