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.
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.