Limiting Reactant
Lab 3
Page 67 to 74 : Lab Notebook
Pre-Lab = Page 71 – All questions
Lab Questions: 1, 2, 3, 5
Page 140: Brady
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Determining Limiting Reactants in Chemical Reactions: A Lab Experiment, Exams of Stoichiometry
Instructions for a lab experiment aimed at determining the limiting reactant in a chemical reaction between sodium phosphate and barium chloride. the concept of limiting reactants, the method for their identification, and the importance of knowing the limiting reactant. The experiment involves the formation of barium phosphate as an insoluble product and the analysis of the supernatant to identify the excess ions. The document also includes background information on the balanced chemical equation and the role of spectator ions.
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2021/2022
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Limiting Reactant
Lab 3
Page 67 to 74 : Lab Notebook Pre-Lab = Page 71 – All questions Lab Questions: 1, 2, 3, 5 Page 140: Brady
What is a limiting reactant?
- It is the reactant that limits the product that
can be formed.
- Stoichiometry allows us to compare the
amounts of various species involved in areaction.
Method for determining limiting reactant 1.
The equation must be balanced.
The amount of two (or more) of thereactants involved, must be known.
Treat each as a separate stoichiometryproblem.
Evaluate the amount of product made or reagentrequired of each unknown species based on theknown quantities. This is determined bycomparing the number of moles of productformed for each mole of the reactant used. Dothis for each of the known reactants.
Compare the amount of product made by eachreagent.
The limiting reagent is the one that produces theleast amount and is the one that any subsequentcalculations should be based on.
Lab Objectives • To determine the limiting reactant in a
mixture of insoluble salts
- Determine the % composition of each
substance in a salt mixture
Background Information for Lab
2Na
PO 3
•12H 4
O 2
(aq)
+3BaCl
•2H 2
O 2
(aq)
Æ
Ba
(PO 3
4
)2(s)
+6NaCl
(aq)
30H
O 2
(l)
Barium Phosphate is the insoluble product Sodium Chloride remains in solution The ionic equation can be written: 3Na
+^
- 2PO
3- 4
- 3Ba
2+
- 2Cl
-^
Æ
Ba
(PO 3
) 4
2 (s)
+6Na
+^
- Cl
- 30H
O 2
The ‘spectator’ ions can be cancelled out, leaving the net ionic eqn.
2PO
3- 4
(aq) + 3Ba
2+
(aq)
Æ
Ba
(PO 3
) 4
2 (s)
Using the balanced equation:
•^
Molar mass of Na
PO^3
•12H 4
O = 380.12 2
•^
The water is included in the formula mass
-^
To calculate the % of each ion present we can usethe molar mass
-^
% of H
O present in formula – 2
- H
O Molar mass = 18.01 2
- 12 are present in the formula = 12 x 18.01 = 216.12 – % of H
O present = 216.12 / 380.12 x 100 = 56.86% 2
2PO
3- 4
(aq) + 3Ba
2+
(aq)
Æ
Ba
(PO 3
) 4 2 (s)
-^
Ratio of reactants is 2:3 (1 : 1.5)
-^
In this experiment, known masses of sodium phosphateand barium chloride will be reacted.
-^
For example:^ – If you have 1g of sodium phosphate and need to
calculate the moles of Barium required
- Calculate the no. of moles of Sodium phosphate – # of moles = mass / Molar mass
0.00263 moles Sodium phosphate
•^
You will be provided with a sample of unknowncomposition of Na
PO 3
•12H 4
O/ BaCl 2
2
•2H
O 2
•^
This will be added to water
-^
A precipitate of barium phosphate will form
-^
This will be collected by filtration, dried andweighed
-^
The supernatant (the liquid left after the solid isremoved) will be analyzed to see what ions are leftin solution
1. measure a mass of solidsodium phosphate/barium chloride mix2. Add water
Ratio of reactants is unknown
3. A precipitate of barium phosphate is formed4. Use filtration to collect the Ba
(PO 3
) 4 2 The Ba
(PO 3
)^42
is recovered fromthe filter paper
supernatant issaved and divided into 2test tubes
test for excessPO
3- 4 ions, add Ba
2+^
ions
test for excessBa
2+^
ions, add PO
3- 4 ions
Formation of a precipitate indicates the presence of that ion.Therefore that ion is present in excess and is not the L.R
Remove the precipitate from thefilter paper, dry and record weight
Using the data to calculate compositionof the salt• Use what you know:
- Weight of initial sample (e.g., 0.942g)– Formula of precipitate: Ba
(PO 3
2
- Molar mass of Ba
(PO 3
2
- Amount of Ba
(PO 3
2
formed, this is the solid
collected by filtration (e.g. 0.118g)
- What was left in solution (excess Barium or
phosphate ions), therefore know the limitingreactant.
Using the mass collected of Ba
(PO 3
, the 2
no of moles can be calculated:^ # of moles of Ba
(PO 3
2
of moles of Ba
(PO 3
2
= 0.00031 (3.12 x
Going back to the balanced ionic equation:
2PO
3- 4
(aq) + 3Ba
2+
(aq)
Æ
Ba
(PO 3
) 4
2
it can be seen that every 1 mole of product
(precipitate) is formed from 3 moles of Ba
2+
and
2 moles of PO
3- 4
- Mass of sodium phosphate can be determined
from the initial mass weighed.
- 0.942g was weighed of the mixture • Mass of Barium Chloride = 0.229g • Mass of
Na
PO 3
•12H 4
O = 0.942 – 0.229 2
- Mass of
Na
PO 3
•12H 4
O = 0.713g 2
•^
% of Na
PO 3
•12H 4
O in sample = 0.713 / 0.942 x 2
•^
% of Na
PO 3
•12H 4
O in sample = 75.7% 2
Lab Techniques
- Gravity Filtration • Filtration is a technique used either to