LIMITING REAGENT PRECIPITATION
OBJECTIVES:
To determine the limiting reagent in the formation of a precipitate.
To determine the composition of a salt mixture.
BACKGROUND:
Two factors that limit the yield of products in a chemical reaction are (1) the amounts of
starting materials (reactants) and (2) the percent yield of the reaction. Many experimental
conditions-for example, temperature and pressure-can be adjusted to increase a reaction's
yield, but because chemicals react according to fixed mole ratios (stoichiometrically), only a
limited amount of product is formed from given amounts of starting material. The reactant
restricting the amount produced is the limiting reagent in that chemical system.
To better understand the limiting reagent, look at the reaction studied in this
experiment. The equation's molecular form is
BaCl2(aq) + Na2SO4(aq) ---> BaSO4(s) + 2 NaCl(aq)
(1)
Since BaCl2 (in the form of the dihydrate, BaCl2.2 H2O) and Na2SO4 are soluble salts
and since BaSO4 is insoluble, the ionic equation is
+
Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO24 (aq) ---> BaSO4(s) + 2 Na (aq) + 2 Cl (aq)
(2)
Cancelling out spectator ions, the net ionic equation is
Ba2+(aq) + SO24 (aq) ---> BaSO4(s)
(3)
One mole of barium ions (from 1 mole of BaCl2.2H2O (244.26 g)), reacts with 1 mole of
sulfate ions (from 1 mole of Na2SO4 (144.04 g)), in the production of 1 mole of barium sulfate
precipitate (233.39 g), if the reaction proceeds to completion.
Suppose, however, only 200 g of BaCl2.2 H2O and 120 g of Na2SO4 are in the reaction
vessel. How many moles and grams of barium sulfate are then produced? Since the equation
reads in terms of moles, not grams, each reactant's number of moles must be determined.
Therefore,
1 mol BaCl2.2H2O
1 mol Ba2+
200 g BaCl2.2H2O
= 0.819 mol Ba2+
244.26 g BaCl2.2H2O
1 mol BaCl .2H O
(
(
1 mol Na2SO4
120 g Na2SO4 144.04 g Na SO
2
4
)(
)(
1 mol SO24
1 mol Na2SO4
2
2
)
) = 0.833 mol SO
24
Equation (3) indicates that 1 mol Ba2+ reacts with 1 mol SO24 . However, since only 0.819 mol
Ba2+ is in the reaction vessel, only 0.819 mol SO2precipitates;
not enough Ba2+ is present for
4
2+ reacts with SO2- in a 1:1 mole ratio, only 0.819 mol
all of the SO24 to precipitate. Since Ba
4
BaSO4 precipitates.
In other words, of the initial 0.833 mol of SO24 in the mixture, only 0.819 mol reacts
2leaving an excess of (0.833 - 0.819 = ) 0.014 mol SO24 . SO 4 is termed the excess reagent.
2+
Because all of the Ba reacts, it is termed the limiting reagent, limiting the amount of BaSO 4
that can be produced.
Once the limiting reagent is known, the theoretical yield of product is calculated from
the balanced equation. Since Ba2+ is the limiting reagent, using the balanced equation (3), we
arrive at the number of moles.
1 mol Ba2+ produces 1 mol BaSO4
or
0.819 mol Ba2+ produces 0.819 mol BaSO4
and
0.819 mol BaSO4
g BaSO
(233.39
) = 191 g BaSO
1 mol BaSO
4
4
4
Thus, 191 g BaSO4 form if the reaction is 100% complete.
In this experiment an unknown mixture of salts, Na2SO4 and BaCl2.2 H2O, is added to
water where according to equation (1) they react forming insoluble BaSO4. Measuring the mass
and calculating the moles of precipitated BaSO4 allows you to calculate the moles and masses
of BaCl2.2 H2O and Na2SO4 that react to form BaSO4.
From a series of tests the limiting and excess reagents are determined. The difference
between the mass of the original salt mixture, mSM, and the mass of the limiting reagent, mLR,
yields the mass of excess reagent, mXR, in the salt mixture.
mXR = mSM - mLR
(4)
By knowing the masses of both the limiting and excess reagents in the original salt mixture,
the percent composition of the salt mixture can be determined. Thus, we can determine the
limiting reagent and the salt mixture's composition.
PROCEDURE:
Two trials are recommended for this experiment. To hasten the analyses, weigh
duplicate unknown salt mixtures, and simultaneously follow the procedure for each. Label the
beakers accordingly for Trial 1 and Trial 2 to avoid intermixing samples and solutions.
Part A. Precipitation of BaSO4
1.
Weigh on weighing paper (+
- 0.001 g) about 1.0 g of an unknown salt mixture.
Transfer it to a 400 mL beaker and add 200 mL of distilled water. Add, using a
stirring rod, 1 mL concentrated HCl (CAUTION: Concentrated HCl is a severe
skin irritant--flush the affected area with a large amount of water). Stir the
aqueous mixture with the stirring rod for about 1 minute and then allow the
precipitate to settle.
2.
Cover the beaker with a watchglass and maintain the solution at 80-90°C with a
low flame for 1 hour. Allow the precipitate to settle.
3.
Decant two 50 mL volumes of the supernatant liquid into separate 100 mL
beakers, labeled Beaker I and Beaker II, and save for Part B.
4.
Filter the BaSO4 precipitate by vacuum filtration. Use fine porosity filter paper
such as Whatman No. 42. Weigh (+
- 0.001 g) the dry filter paper and seal it into
the funnel with a few milliliters of distilled water. Discard this water from the
receiving flask.
5.
While the solution is still hot, quantitatively transfer the precipitate to the
funnel. Remove any precipitate on the beaker's wall with a rubber policeman and
with hot water wash the precipitate onto the filter. Rinse the precipitate on the
filter paper with two 5 mL portions of hot water.
6.
First, dry the precipitate on filter paper, then place it in your desk drawer until
later. Weigh (+
- 0.001 g) the filter paper and precipitate. Record.
Part B. Determination of Limiting Reagent
1.
Testing for excess Ba2+. Add 2 drops of 0.5 M SO24 ion (from 0.5 M Na2SO4) to
the 50 mL of solution in Beaker I. If a precipitate forms, Ba2+ is in excess and
SO24 is the limiting reagent.
2+ ion (from 0.5 M BaCl ) to the
2.
Testing for excess SO22
4 . Add 2 drops of 0.5 M Ba
other 50 mL of solution in Beaker II. If a precipitate forms, SO 2is
in
excess
and
4
2+
Ba is the limiting reagent.