Does Any Chemistry Professor that Knows THis Question?

Question

Hi,
i did this experiment and when i calculated the formula or missing compund it doesnt seem right.
can you please help; i have listed the experiment, the given & found values, and what to find below.

Thank you in advance.
------------------------------------------------------------------
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(-) + ?

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.

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(-)

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.

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

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.


--------------------------

Please answer only if you know...
Dont reply if you have nothing good to say...

Answer 1

You'd be better off asking your question on a chemistry web page like Scientific American:

Answer 2

ok. here we go:

QUESTION A:

number of moles of KBrO3 initially delivered
from the formula,
conc (M) = no. mol (mol) / volume (L),
you get no. mol (mol) = conc (M) x volume (L)
= 0.020M x 0.020L
= 0.00040mol

QUESTION B:

Number of moles of KBrO3 remaining in solution:

titration 1 - moles Na2S2O3 used = 0.100M x 0.0132L
= 0.00132 mol
moles S2O3 used = moles Na2S2O3
= 0.00132mol
from equation, moles I3- used = 1/2 x moles S2O3
= 1/2 x 0.00132
= 0.000660mol
from equation, moles BrO3- = 1/3 x moles I3-
= 1/3 x .000660mol
= 0.000220mol
moles KBrO3 = moles BrO3-
= 0.000220mol

titration 2 - moles Na2S2O3 used = 0.100M x 0.0130L
= 0.00130 mol
moles S2O3 used = moles Na2S2O3
= 0.00130mol
from equation, moles I3- used = 1/2 x moles S2O3
= 1/2 x 0.00130
= 0.000650mol
from equation, moles BrO3- = 1/3 x moles I3-
= 1/3 x .000650mol
= 0.000216mol
moles KBrO3 = moles BrO3-
= 0.000216mol

titration 3 - moles Na2S2O3 used = 0.100M x 0.0141L
= 0.00141 mol
moles S2O3 used = moles Na2S2O3
= 0.00141mol
from equation, moles I3- used = 1/2 x moles S2O3
= 1/2 x 0.00141
= 0.000705mol
from equation, moles BrO3- = 1/3 x moles I3-
= 1/3 x .000705mol
= 0.000235mol
moles KBrO3 = moles BrO3-
= 0.000235mol

QUESTION C:

number of moles of KBrO3 that reacted:

titraltion 1: moles = 0.00040mol - 0.000220mol
= 0.000180mol

titraltion 2: moles = 0.00040mol - 0.000216mol
= 0.000184mol

titraltion 3: moles = 0.00040mol - 0.000235mol
= 0.000165mol

QUESTION D:

average moles KBrO3 reacting

average = (0.000180 + 0.000184 + 0.000165) / 3
= 0.000176mol

QUESTION E:

moles electrons transferred between BrO3-

the oxidation state of Bromine in BrO3- is +5, it's oxidation state in Br- is -1. This means that each bromine atom MUST gain 6 electrons during the reaction.

moles electrons transferred = 5 x moles BrO3-
= 6 x 0.000176mol
= 0.001056mol

QUESTION F:

change in oxidation state of N:

moles hydroxylammonium = conc (M) x volume (L)
= 0.020M x 0.010L
= 0.00020mol

change = moles e- transferred / moles hydroxylammonium
= 0.001056mol / 0.00020mol
= 5.28
= 5

QUESTION G:

the two that match in the list of possible products (from your results, where the N has on oxidation state of 5) are NH4+ and NO3-. Your lab says it's in acidic solution, so it's probably NO3-.


If this i correct, you OWE ME, big time. It took a while to work it out.


Hope this helps.