# Stoichiometry and Limiting Reagent

Topics: Stoichiometry, Sodium, Reagent Pages: 5 (1141 words) Published: October 19, 2011
Stoichiometry and Limiting Reagents
Theodore A. Bieniosek

I. Purpose and Theory

The purpose of the experiment is to study and apply the processes of stoichiometric calculation on a controlled chemical reaction. We will be adding variable amounts of reactants in a chemical reaction in order to demonstrate the effect of limiting reagents. Based on the volumes of the reactants, and their respective molarities, we can calculate the theoretical yield of the reaction and compare it to the amount of products experimentally yielded. From the molarities of the solutions, the amount of moles can be calculated, and in turn fed into the balanced chemical equation to determine the theoretical yield.

II. Data and Calculations

Data
Test #1Test #2Test #3
Initial Volume of Sodium Hydroxide:42.4 ml26.6 ml30.6 ml
Final Volume of Sodium Hydroxide:44.4 ml30.6 ml35.6 ml
Volume of Sodium Hydroxide Delivered:2.0 ml4.0 ml5.0 ml

Initial Volume of Calcium Chloride:42.6 ml39.45 ml10.7 ml Final Volume of Calcium Chloride:47.6 ml44.45 ml15.7 ml
Volume of Calcium Chloride Delivered:5.0 ml5.0 ml5.0 ml

Excess Reagent Test
with Sodium Hydroxide:positivepositivenegative
with Calcium Chloride:negativenegativepositive

Initial Mass of Filter Paper:.89 g.87 g.76 g
Final Mass of Filter Paper
with precipitate:1.14 g1.27 g1.15 g
Mass of Precipitate.25 g.40 g.39 g

Calculations

CaCl2(aq) + 2NaOH(aq) Ca(OH)2(s) + 2NaCl(aq)

Moles = [volume (L)] * (Molarity)

Molarity CaCl2: 1.0 M
Molarity NaOH: 2.5 M

Test #1
Volume CaCl2: 5.0 mL = .005 L
Volume NaOH: 2.0 mL = .002 L

Moles CaCl2 = (.005 L) * (1.0 mol/L) = .005 moles
Moles NaOH = (.002 L) * (2.5 mol/L) = .005 moles

Limiting Reactant:
NaOH, because precipitate was formed when Sodium Hydroxide was added to filtered solution, CaCl2 was in excess, meaning the NaOH was completely used up and therefore limited the reaction.

[ (2 mole NaOH) / (1 mole CaCl2) ]*[ .005 mol CaCl2 ] = .010 moles NaOH needed to completely react CaCl2, only .005 moles available, therefore NaOH limits the reaction

Theoretical Yield of Ca(OH)2:
[ .005 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = .0025 moles Ca(OH)2 formed
[ .0025 moles Ca(OH)2 ] * [ (74.0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = .1852 grams Ca(OH)2

Percent Yield
[ .25 g / .1852 g ] * 100 % = 134.989 %

Test #2

Volume CaCl2: 5.0 mL = .005 L
Volume NaOH: 4.0 mL = .004 L

Moles CaCl2 = (.005 L) * (1.0 mol/L) = .005 moles
Moles NaOH = (.004 L) * (2.5 mol/L) = .010 moles

Limiting Reactant:
NaOH, because precipitate was formed when Sodium Hydroxide was added to filtered solution, CaCl2 was in excess, meaning the NaOH was completely used up and therefore limited the reaction.

[ (2 mole NaOH) / (1 mole CaCl2) ]*[ .005 mol CaCl2 ] = .010 moles NaOH needed to completely react CaCl2, only .010 moles available, theoretically all reagents are completely reacted. Measurement error has apparently left excess CaCl2 in filtered solution, and lack of NaOH limited the reaction.

Theoretical Yield of Ca(OH)2:
[ .010 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = .005 moles Ca(OH)2 formed
[ .005 moles Ca(OH)2 ] * [ (74.0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = .3705 grams Ca(OH)2

Percent Yield
[ actual yield / theoretical yield ] * 100%
[ .40 g / .3705 g ] * 100% = 107.962 %

Test #3

Volume CaCl2: 5.0 mL = .005 L
Volume NaOH: 5.0 mL = .005 L

Moles CaCl2 = (.005 L) * (1.0 mol/L) = .005 moles
Moles NaOH = (.005 L) * (2.5 mol/L) = .0125 moles

Limiting Reactant:
CaCl2, because precipitate was formed when Calcium Chloride was added to filtered solution, NaOH was in excess, meaning the CaCl2 was completely used up and therefore limited the reaction.

[ (1 mole CaCl2) / (2 mole NaOH)...

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