Question

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 weighed. What is the mass, in g, of the precipitated silver bromate? Assume silver bromate is completely insoluble.

I've calculated the moles and have this equation so I know it is a 1 to 1 ratio...but I'm completely stumped now:

KBrO3 + AgNO3 ----> KNO3 + AgBrO3

Answer 1

You have to calculate the moles of the potassium bromate and silver nitrate from the masses.

Then, take the LESSOR of these as the limiting reactant. There is a 1:1 ratio, so you get the same number of moles of silver bromate. Convert to grams.

Answer 2

That means that AgNO3 is the limiting reactant as it has 0.0033 moles compared to 0.0039 for silver nitrate.

But I don't get the last part you said. What same number of moles of silver bromate would I get? Do you mean same number of moles as the limiting reactant? Meaning that there would be 0.0033 moles of silver bromate?