Finding the precipitate of the reaction of KBr with AgNO3

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SUMMARY

The discussion focuses on calculating the mass of precipitated silver bromate (AgBrO3) from a reaction between potassium bromate (KBr) and silver nitrate (AgNO3). The participants confirm that the volume of water (388 mL) does not affect the mass calculation of the precipitate, as it is assumed that silver bromate is completely insoluble. The key steps involve determining the moles of each reactant using the formula n=m/M, identifying the limiting reagent, and applying the molar ratio to find the mass of the precipitate.

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  • Understanding of stoichiometry and limiting reagents
  • Familiarity with the concept of molar mass (M)
  • Knowledge of precipitation reactions in chemistry
  • Ability to perform calculations involving moles (n) and mass (m)
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  • Calculate the molar mass of silver bromate (AgBrO3)
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  • Study the concept of limiting reagents in chemical reactions
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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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