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

The chemical reaction can be expressed with this balanced equation:
 <span>SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s) 
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Using the periodic table:
Silver has a mass of 108 grams
Bromine weighs 80 grams

Thus, the molar mass of silver bromide amounts to 188 grams (108 + 80).

To find the number of moles, we use the formula: number of moles = mass / molar mass
The moles of precipitate formed can be calculated as 3.491/188 = 0.018 moles.

According to the balanced equation:
1 mole of strontium bromide gives rise to 2 moles of silver bromide. Therefore, we can find the moles of strontium bromide required to create 0.018 moles of silver bromide by cross-multiplying as follows:
amount of<span>strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles</span>