For each of the following precipitation reactions calculate how many grams of the first reactant are necessary to completely react with 55.8 g of the second reactant. K2SO4 + Sr(NO3)2 + SrSO4 + 2KNO3

Answer 1

I think you mean
K2SO4(aq) + Sr(NO3)2(aq) > SrSO4 (s)+ 2KNO3 (aq)

so you need one mole of the first reactant for each mole of the second reactant.
How many moles of strontium nitrtate=55.8/211.6
so you need that number of moles of postassium sulfate, and in grams that is
55.8/211.6 * formulamassK2SO4=
55.8/211.6 *174.3 grams