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5. a reaction occurs between solutions of strontium bromide and silver nitrate, as shown in the equation below: srbr2 (aq) + AgNO3 (aq) → Sr(NO3)2 (aq) + AgBr (s) a. if 3.491 grams of the precipitate is formed, how many moles of strontium bromide were reacted?
Accepted Answer
To determine the moles of strontium bromide reacted, we need to follow these steps:
1. Find the moles of AgBr precipitate:
- Divide the mass of AgBr (3.491 g) by its molar mass (187.77 g/mol):
- Moles of AgBr = 3.491 g / 187.77 g/mol = 0.0186 mol
2. Use the stoichiometry of the balanced equation:
- The balanced equation shows a 1:1 mole ratio between SrBr2 and AgBr.
- This means 1 mole of SrBr2 reacts to form 1 mole of AgBr.
3. Calculate the moles of SrBr2:
- Since the mole ratio is 1:1, the moles of SrBr2 reacted are equal to the moles of AgBr formed.
- Moles of SrBr2 = 0.0186 mol
Therefore, 0.0186 moles of strontium bromide were reacted.
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