Finding the precipitate of the reaction of KBr with AgNO3

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The discussion centers on calculating the mass of precipitated silver bromate from the reaction between potassium bromate and silver nitrate. The main concern is whether the volume of water (388 mL) affects the calculations, with participants suggesting it may not be relevant for determining the precipitate mass. The focus is on finding the moles of each reagent to identify the limiting reagent, which is essential for calculating the mass of the precipitate using the molar ratio. Clarification is sought on the role of water in the context of this specific problem. Ultimately, the volume of water does not impact the calculation of the precipitate mass directly.
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Homework Statement


0.658 g of potassium bromate and 0.562 g of silver nitrate are added to 388 mL water. Solid silver bromate is formed, dried, and weighted. What is the mass, in g, of the precipitated silver bromate? Assume silver bromate is completely insoluble.


Homework Equations


n=m/M


The Attempt at a Solution


The only actual problem I have with this question is whether or not the 388 mL water comes into play with the calculations. If it doesn't, then I pretty much just find the moles of each reagent and find the limiting reagent, which can then be used to find the mass of the precipitate based on the molar ratio. But if I somehow need to incorporate the water variable into the calculations, then I'm a bit confused.
 
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I can't see how the amount of water comes in - other than that your method is fine.
Sometimes the amount of water is there because this is/was part of a longer question where you had to work out something else as well.
 
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