Determine the Limiting Reagent
Investigation 3 Limiting Reagents
Aim: To determine the limiting reagent in a reaction.
Hypothesis: if the number of moles of the solutions added together equals the stoichiometry ratio in the balanced equation, then we would not expect a limiting reagent in the reaction.
Independent variable: The number of moles of the reactants
Dependent Variable: The limiting reagent
Constant Variable: The volume of reactants added
2 100 ml beaker
2 measuring cylinders
10 ml of 0.1 molL-1 HCL(aq), NAOH(aq)
20 ml of 0.1 molL-1H2SO4(aq), CuSO4(aq)
30 ml of 0.1 molL-1 Ba(NO3)2(aq)
4 test tubes
1) Measure 10 ml of 0.1 molL-1 of Hydrochloric Acid using a measuring cylinder. 2) Measure 10 ml of 0.1 molL-1 of Sodium Hydroxide solution using another measuring cylinder. 3) Pour the 2 solutions into a 100 ml beaker.
4) Mix the solution using a stirring rod.
5) Add 2 drops of the universal indicator into the solution. 6) Observe any changes to the resulting solution. (colour)
The solution turns light purple after adding the universal indicator.
1) Measure 10 ml of Ba(NO3)2(aq) and H2SO4(aq).
2) Pour the 2 solutions into a 100 ml beaker.
3) Mix using a stirring rod.
4) Prepare a funnel on top of a 100 ml beaker.
5) Cover the top of the funnel with 3 layers of filter paper. 6) Pour the solution into the funnel.
7) Separate the solution obtained in the beaker into 2 test tubes. 8) Add Ba(NO3)2 into one of the test tubes and H2SO4 into the other. 9) Observe and record any changes.
Repeat the method used in Reaction B using Ba(NO3)2 and CuSO4 instead.
Test tube 1 | Ba(NO3)2 added to the solution from beaker | Precipitate formed| Test tube 2| H2SO4 added to the solution from beaker| No change|
Test tube 1| Ba(NO3)2 added to the solution from beaker| Precipitate formed| Test tube 2| CuSO4 added to the solution from beaker| No change|
Processing of data:
In Reaction A, the resulting solution of HCl and NaOH turns light purple after adding 2 drops of the Universal Indicator. Based on the colour chart, it shows that this solution has a pH value of approximately 10, which means it is basic. The equation is :
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
HCl is an acid whereas NaOH is basic. The pH value of 10 means that the reaction is not complete as it contains excess NaOH(a base), giving it its basic property. Hence, we can conclude that HCl is the limiting reagent.
In Reaction B and C, a different method was used to determine the limiting reagent of a reaction.
The equation of the reaction when Ba(NO3)2 and H2SO4 and mixed is:
Ba(NO3)2(aq) + H2SO4(aq) BaSO4(s) + 2HNO3(aq)
Hence the solution in the beaker collected would not be BaSO4, as the precipitate would have been collected on the filter paper. By adding the solution collected in the beaker back to the 2 separate initial solutions, a precipitate is formed in the test tube containing Ba(NO3)2.
This means that the reaction in test tube 1 has gone as below:
HNO3 + H2SO4 + Ba(NO3)2 BaSO4(s) + 3HNO3(aq)
Ionic eqn – SO42- + Ba2+ BaSO4(s)
The precipitate formed is BaSO4, which means that there is excess H2SO4 in the solution in the beaker. The solution is expected to only contain HNO3 if the initial reaction had gone to completion, however the presence of excess H2SO4 results in a precipitate forming when added to Ba(NO3)2, based on the equation above.
In test tube 2, the equation is:
HNO3 + H2SO4 + H2SO4 2H2SO4 + HNO3
There is no reaction, hence no change in the resulting solution.
This proves that H2SO4 is present in excess, making Ba(NO3)2 the limiting reagent in this reaction.
As for Reaction C,...