If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted

The balanced chemical equation that represents the reaction is as follows:
 SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s) 

From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams

molar mass of silver bromide = 108 + 80 = 188 grams

number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles

From the balanced equation:
1 mole of  strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles