what mass of solid AgBr is produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0mL of 1.00 MNaBr?

Question

i keep getting the answer 3.75 g however when i checked in the back of the book it gives me 2.82 g.. can you please explain the universal steps for me

Answer 1

The universal steps:
1) balanced equation
AgNO3 + NaBr >> NaNO3 (aq) + AgBr (s)

balanced.
So, one mole of AgBr is produced for each mole of silver nitrate and sodium bromide.

2) number moles you started with:
AgNO3= .1*.150= .0150moles

NaBr= .020*1= .0200 moles

3) determining the limiting reactant: Since the reactants react on a one to one basis, the AgNO3 is the limiting reactant. You have .0150 moles of that.

4) determine the moles of product:
.0150 moles of AgBr, from the balanced equation

5) mass produced: .0150*molmassAgBr=.0150*188=2.82 grams

Answer 2

2.82g