Calculate the concentration of ammonium ions and sulfate ions in the final solutio?

find moles, using molar mass:
17.9 g (NH4)2SO4 @ 132.14 g/mol = 0.13546 moles (NH4)2SO4

find molarity:
0.13546 moles (NH4)2SO4 / 0.1000 litres = 1.3546 Molar  (NH4)2SO4

12.00 ml of 1.3546 Molar  (NH4)2SO4 gets diluted by adding 60.00 ml :
C1V1 = C2V2
(1.3546 M)(12.00ml) = C2(72.00)
C2 = 0.2258 Molar NH4)2SO4

0.2258 Molar NH4)2SO4 releases 100% of its ions:
0.2258 Molar (SO4)-2 ion, @ 1:1 ratio
0.4515 Molar (HN4)+ ion , @ 1:2 ratio

your answers(3 sigfigs because of "17.9" grams):
0.226 Molar (SO4)-2 ion,
0.452 Molar (HN4)+ ion


( if the ammonium sulfate was actually "17.90 grams, it would be a 4 sigfig calc)