Calculating Mass of Precipitated Silver Bromate - Chemistry Homework

[a.a]

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?

 

[GCT]

Method seems fine , the significant figures should be 3.

 

[Borek]

K_so = 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...

 

[Pete5x5]

That method worked for me as well. Answer would be 6.68e-1