Since Sr(NO3)2 is soluble in water, it will be in its ionic form Sr2+ + NO3-. H2SO4 is a diprotic acid, and it's hard for it to lose the second H, but since SrSO4 precipitates out of the solution, it drives the equilibrium towards the products. So in the end you'll have Sr2+(aq) + SO4 2-(aq) -> SrSO4 (s)