gogeta2006
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Determining the Product of Redox reaction
Hello Guys,
My name is Mani. I am a first year undergraduate student and i need help. In the experiment listed below we have to determine the product of redox
reaction.
I have listed completely the lab instructions and is not that i cannot do the lab, im not getting the correct results. Please if you can take some
time, read the lab and do the questions.
It may seem very long but the lab is very short, even the calculations in the end (questions). Can someone please do this and tell me the results with
explanation.
Thank you soo much for your consideration and generosity.
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Experiment:
In this experiment you will determine the equation of a redox reaction by the use of experimental data.
This is a reaction in which it is not obvious what products will be formed but by finding the relative number of moles reacting
you can deduce what reaction has occured.
The reaction is between bromate ions, BrO3(-) and hydroxylammonium ions NH3OH(+) in aqueous acid solution.
You assume that the bromate are converted to bromide ions, Br(-) (How can this reaction be shoed experimentally)
So the reaction that is being studied is:
BrO3(-) + NH3OH(+) ----> Br(-) + ?
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If you know the number of moles of bromate consumed in the reaction and you know the oxidation states of Br in the reactant and in the product, then
you can calculate the number of moles of electrons gained by Br in the reaction. (this mist equal to the number of moles of electrons lost by the N of
hydroxylammonium in the reaction. Knowing the number of moles of hydroxylammonium reacted, one can calculate the number of moles lost by N per
hydoxylammonium reacted. This then allows you to determine the oxidation state of the N in the unknown product. In the acid solution, the following
compounds of N and H and/or O can exist: NH4(+), N2H5(+), HN3, N2, N2O, NO, H2N2O3, HNO2, No2, No3(-).
It can be assumed that the N containing product of the reaction is one of these.
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The experiment is arranged so that the hydoxylammonium is the limiting reagent and bromate is in excess. The amoun tof this excess is found by adding
potassium iodide solution, which reacts with bromate:
BrO3(-) + 9I(-) + 6H3O(+) ---> Br(-) + 3I3(-) + 9H20
You can assume that nothing else reacts with the iodide ions. The resulting triioxide ions, I3(-), are brown in colour and can be
titrated with thiosulphate, S2O3^2(-):
I3(-) + 2S2O3^2(-) ---> 3I- + S4O6^2(-)
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From the titration the number of moles of triiodide is determined. From the number of moles of triiodide and the two balanced equations above, the
number of moles of excess bromate is determined. Subtracting the number of moles of excess bromate from the total number of moles of bromate used,
gives us the number of moles of bromate consumed in the reaction. From this the N containing product can be identified and a balanced equation
for this reaction can be written.
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Given and Calculated:
Nh3OHCl:
concentration = 0.020 M
Volume used in each reaction = 10mL
KBrO3:
concentration = 0.020 M
volume used in each reaction = 20 mL
Na2S2O3:
concentration: 0.100M
flask 1:
inital burette reading: 0 mL
final burette reading: 13.2 mL
flask 2:
inital burette reading: 13.2 mL
final burette reading: 26.2 mL
flask 3:
inital burette reading: 26.2 mL
final burette reading: 40.3 mL
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Questions:
A) For each of the three sets of data calculate the NUMBER OF MOLES of KBrO3 initially delivered, taking into account its volume of KBrO3 delievered
and its concentration.
B) From the result of the titration of unreacted KBrO3 with Na2S2O3, find the NUMBER OF MOLES of KBrO3 remaining in the solution ( there was a 10
minute reaction period - not relevent - i think).
C) FIND THE DIFFERENCE between the two values (answers of a and b which will give you the number of moles of KBrO3 which reacted with the
hydroxylammonium ion (NH3OHCl).
D) CALCULATE THE AVERAGE of three values for the moles of KBrO3 which reacted
.
E) FIND THE NUMBER OF MOLES OF ELECTRONS which must have been transferred between BrO3(-) and NH3OH(+) during this reaction.
In order to do this you take into account the change in oxidation state which the Br undegoes when bromate reacts to become bromide.
F) Divide number of moles of electrons transferred by the number of moles of hydroxylammonium ions reacted to find the number of moles of electrons
lost by each mole of hydroxylammonium ion.
This is equal to the change in the oxidation state of N during reaction.
G) From your knowledge of the oxidation state of nitrogen in the product, as determined experimentally, and the oxidation state of nitrogen in each
species in the list of possible products,
select the most likely product.
WRITE THE BALANCED EQUATION for the reaction between bromate and hydroxylammonium ions in acid solution
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guy
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It probably was reduced to nitrogen gas. Did you see any gas?
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gogeta2006
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No n0t really just change in color
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