Finding the precipitate of the reaction of KBr with AgNO3

[MiniJo]

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.

 

[mgb_phys]

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.

 

[Blueshift5]

I would approach this problem by first identifying the reactants and products of the reaction. In this case, the reactants are potassium bromate (KBr) and silver nitrate (AgNO3), and the product is silver bromate (AgBrO3).

Next, I would calculate the moles of each reactant by dividing the given mass by its molar mass. This would give me 0.0063 moles of KBr and 0.0035 moles of AgNO3.

Then, I would use the mole ratio between KBr and AgBrO3 to determine the limiting reagent. In this case, the mole ratio is 1:1, meaning that both reactants are present in equal amounts. This means that both reactants will be completely used up in the reaction and no excess will be left over.

Since silver bromate is completely insoluble, it will precipitate out of the solution. This means that the mass of the precipitate will be equal to the mass of the product formed, which is 0.0063 moles of AgBrO3.

To calculate the mass of the precipitate, I would use the formula n=m/M, where n is the number of moles, m is the mass, and M is the molar mass. Plugging in the values, I would get a mass of 0.724 g of silver bromate.

Therefore, the mass of the precipitate of the reaction of KBr with AgNO3 is 0.724 g. This mass can be verified by weighing the solid silver bromate after it is dried.