Calculating Mass of Precipitated Silver Bromate - Chemistry Homework

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To calculate the mass of precipitated silver bromate (AgBrO3) from the reaction of silver nitrate (AgNO3) and potassium bromate (KBrO3), the limiting reactant is determined to be KBrO3. Given 0.790 g of AgNO3 and 0.473 g of KBrO3, the calculations yield 1.0965 g of AgBrO3 from AgNO3 and 0.668 g from KBrO3. Therefore, the mass of the precipitated silver bromate is 0.668 g. The significant figures for the final answer should be three, resulting in 6.68e-1 g. The discussion also touches on the solubility of AgBrO3 in the context of the calculations.
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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?
 
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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
 
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