Mass of silver bromate precipitate from 0.790 g AgNO3 and 0.473 g KBrO3

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Homework Statement


0.790 g of silver nitrate and 0.473 g of potassium bromate are added to 379 mL water. Solid silver bromate is formed, dried, and weighed. What is the mas in g of the precipitated silver bromate?


Homework Equations


AgNO3 + KBrO3 --> AgBrO3 + KNO3

AgNO3 = 169.87 g
KBrO3 = 197.00 g
AgBrO3 = 235.776 g
KNO3 = 101.10 g


The Attempt at a Solution



169.87g AgNO3 => 235.78g AgBrO3
0.790g AgNO3 => (235.78g AgBrO3/ 169.87g AgNO3) * 0.790g AgNO3= 1.0965 g AgBrO3

167.005 KBrO3 => 235.78g AgBrO3
0.473g KBrO3=> (235.78g AgBrO3/167.005 KBrO3)*0.473g KBrO3 = 0.668 g AgBrO3

Limiting Reactant is KBrO3, thus the answer would be 0.668 g AgBrO3, also how many significant figures should we use?
 

Discussion

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Method seems fine , the significant figures should be 3.
 
Kso = 10-4.3 or something close to that number. That means that after 0.214g of AgBrO3 precipitates solution becomes saturated.

Unless what you wrote on chemicalforums is true, and you have to assume AgBrO3 is completely insoluble...
 
That method worked for me as well. Answer would be 6.68e-1