Question

A 38.0 mL sample of 1.10 M KBr and a 57.0 mL sample of 0.570 M KBr are mixed. The solution is then heated to evaporate water until the total volume is 75.0 mL .How many grams of silver nitrate are required to precipitate out silver bromide in the final solution?

Answer 1

1) List the known and unknown quantities.

Solution 1: KBr.

Molarity: 1.10 KBr.

Volume: 38.0 mL = 0.038 L.

Solution 2: KBr.

Molarity: 0.570 M.

Volume: 57.0 mL = 0.057 L.

Total volume: 75.0 mL = 0.075 L.

Molarity: unknown.

2) Molarity of KBr after mixing the two solutions.

2.1-Moles of KBr from solution 1

`mol\text{ }KBr=1.10\text{ }M\text{ }KBr*0.038\text{ }L=0.0418\text{ }mol\text{ }KBr`

2.2-Moles of KBr from solution 2

`mol\text{ }KBr=0.570\text{ }M\text{ }KBr*0.057\text{ }L=0.03249\text{ }mol\text{ }KBr`

Total moles of KBr: 0.0418 + 0.03249 = 0.07429 mol KBr.

2.3-New solution

`M=\frac{0.0418+0.03249}{0.075\text{ }L}=0.9905\text{ }M\text{ }KBr`

3) Moles of KBr needed to react to silver nitrate.

3.1-Write the chemical equation and balance it

`KBr+AgNO_3\rightarrow AgBr+KNO_3`

3.2-Moles of silver nitrate.

The molar ratio between KBr and AgNO3 is 1 mol KBr: 1 mol AgNO3.

`mol\text{ }AgNO_3=0.07429\text{ }KBr*\frac{1\text{ }mol\text{ }AgNO_3}{1\text{ }mol\text{ }KBr}=0.07429\text{ }mol\text{ }AgNO_3`

4) Mass of silver nitrate.

The molar mass of AgNO3 is 169.8731 g/mol.

`g\text{ }AgNO_3=0.07429\text{ }mol\text{ }AgNO_3*\frac{169.8731\text{ }g\text{ }AgNO_3}{1\text{ }mol\text{ }AgNO_3}=12.62\text{ }g\text{ }AgNO_3`

The mass of silver nitrate (AgNO3) required to precipitate out silver bromide is 12.62 g AgNO3.

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