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