Chemistry - stoichiometry, solutions

[excel000]

Homework Statement

0.800 g of silver nitrate and 0.473 g of potassium bromate are added to 238 mL water. Solid silver bromate is formed, dried, and weighed. What is the mass, in g, of the precipitated silver bromate? (Be careful to enter the correct number of significant figures. Do not enter units.) Assume silver bromate is completely insoluble.

Homework Equations

AgNO₃ + KBrO₃ -> AgBrO₃(s) + KNO₃

The Attempt at a Solution

0.800gAgNO3/169.66gmol^-1AgNO3 = 4.7x10^-3mol

0.473gKBrO3/166.998gmol^-1KBrO3 = 2.83x10^-3mol <---- Limiting Reactant

2.83x10^-3molAgBrO3 x 235.77gmol^-1
= 0.667g AgBrO₃

that was my attempt, but i completely ignored the water part since I don't really know what that is for, so I'm pretty sure I'm wrong. any help on this would be greatly appreciated

 

[Borek]

You are right to ignore water, it doesn't take part in the reaction. That's assuming AgBrO₃ is completely insoluble.

https://www.physicsforums.com/showthread.php?t=257528

 

[Blueshift5]

!



Your approach to solving this problem is correct. However, you should also consider the water in the reaction. Since the water is the solvent, it will not participate in the reaction and will not affect the amount of product formed. Therefore, you can still use the given amounts of silver nitrate and potassium bromate to calculate the mass of the precipitated silver bromate, as you have done. However, to determine the final mass of the silver bromate, you should subtract the mass of water used from the calculated mass. This will give you the mass of the silver bromate precipitate alone. Remember to use the appropriate number of significant figures in your final answer. Good job on considering the limiting reactant and using the correct molar masses in your calculations. Keep up the good work!