Question

what mass of silver chloride could be obtained when a solution containing excess potassium chloride is added to 25ml of 2M silver nitrate solution

Answer 1

Answer:

T-27

Tutorial 4

SOLUTION STOICHIOMETRY

Solution stoichiometry calculations involve chemical reactions taking place in solution.

Of the various methods of expressing solution concentration the most convenient for general

laboratory use is molarity, which is defined:

Moles of solute nsolute

Molarity = or M =

Liters of solution Lsoln

Chemical reactions are written in terms of moles of reactants and products; this molarity

concentration unit relates moles of solute to volume of solution. Thus, easily measured solution

volumes provide a simple method of measuring moles of reactants.

EXAMPLE: What is the molarity of a solution made by dissolving 5.67 g of potassium chloride

in enough water to make 100.0 mL of solution?

This data gives a relationship between amount of solute and volume of solution: 5.67 g KCl /

100.0 mL. To find molarity we must convert grams KCl to moles KCl and mL solution to L:

5.67 g KCl 1 mol KCl 1000 mL 0.760 mol KCl

x x = or 0.760 M KCl

100.0 mL 74.6 g KCl L L

Whenever the solution concentration is given in molarity, M, you must change to the equivalent

units, mol/L or mol/1000 mL, to use as a conversion factor.

EXAMPLE: What mass of solute is contained in 15.0 mL of a 0.760 M KCl solution?

The conversion sequence is:

mL solution → L solution → mol KCl → g KCl

1 L 0.760 mol KCl 74.6 g KCl

15.0 mL x x x = 0.850 g KCl

1000 mL L mol KCl

T-28

OR, use the conversion sequence:

mL solution → mol KCl → g KCl

0.760 mol KCl 74.6 g KCl

15.0 mL x x = 0.850 g KCl

1000 mL mol KCl

EXAMPLE: What mass of potassium chloride would be needed to prepare 250.0 mL of a 0.500

M solution?

1 L 0.500 mol KCl 74.6 g KCl

250.0 mL x x x = 9.32 g KCl (needed)

1000 mL L mol KCl

OR

0.500 mol KCl 74.6 g KCl

250.0 mL x x = 9.32 g KCl (needed)

1000 mL mol KCl

Please note: the preceding two EXAMPLES are the same kind of problem worded differently:

conversion from volume of solution to mass of solute.

EXAMPLE: Silver nitrate solution is added to 25.00 mL of a 0.500 M potassium chloride

solution until no more precipitate forms. What mass of silver chloride will be formed?

The chemical equation for the reaction is:

KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq)

Since the concentration and volume of silver nitrate solution are not specified, we can assume it

is in excess. First find the number of moles of KCl in the 25.00 mL of 0.500 M solution:

0.500 mol KCl

(A) 25.00 mL x = 0.0125 mol KCl

1000 mL