Suppose 0.206g of iron(II) bromide is liquified in 100.mL of a 11.0mM aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You have the right to assume the volume of the solution doesn"t readjust when the iron(II) bromide is liquified in it. Be sure your answer has actually the correct number of significant digits. 


Write the appropriately balanced equation:

Find limiting reactant by seeing just how a lot AgBr(s) is formed from each:

For FeBr2: 0.206 g x 1 mole/215.7 g x 2 mole AgBr/1 mole FeBr2 = 0.00191 moles

For AgNO3: 100. ml x 1 L/1000 ml x 11.0 mol/L x 2 mol AgBr/2 mol AgNO3 = 1.10 moles

So, FeBr2 is limiting

Since AgBr is insoluble, there will be no, or exceptionally little bit bromide anion left in solution. It will all be precipitated in the develop of AgBr. If you want to uncover the "actual" concentration of Br- in solution, you need the Ksp value for AgBr.

You are watching: Calculate the final molarity of bromide anion in the solution

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Barbara P.

what if AgNO3 was the limiting reactant?


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