Answer:
0.196M Br^-
Step-by-step explanation:
We have 2.52 L of an aqueous NaBr solution. We add an excess of AgNO3. We form 9.29 gof AgBr(s). What was the initial concentration of Br- (in mol/L) in this aqueous NaBr solution?
We need to start with a balanced equation.
NaBr + AgNO3 = AgBr + NaNO3
This double replacement reaction appears balanced. It tells us the 1 mole of NaBr will react with one mole of AgNO3 to form one mole each of AgBr and NaNO3,
9.29 of AgBr(s) is formed after adding an excess of AgNO3. This means the reaction stopped when all of the NaBr was consumed.
Lets find the moles of AgBr in 9.29 g of the substance. The molar mass of AgBr is 187.8 grams/mole)
(9.29 g)(187.8 grams/mole) = 0.0495 moles AgBr.
The balanced equation tells us that we'll get 1 mole of AgBr for every 1 mole of NaBr, and since we know that excess AgNO3 was added, this means that we started with 0.0495 moles of NaBr. That was dissolved in 2.52 liters of solution. The unit of concentration is Molar, which is defined as moles/liter.
We have 0.0495 moles of NaBr in 2.52 liters, so we can write:
(0.0495 moles AgBr)/(2.52 L) = 0.196M NaBr
Assuming all of the NaBr is dissolved, there would be an equal number of moles of both Na^+ and Br^- (0.196M for each)