Need help with the calculations for a lab data sheet. The lab consists of burnin
ID: 828966 • Letter: N
Question
Need help with the calculations for a lab data sheet. The lab consists of burning hydrogen gas over an ice cube to trap intermediate products, in this case H2O2. I am confused on how to Solve the four questions below using the data collected in experimentation. Thank You!
Detecting a Reaction Intermediate
We can trap H2O2 from the flame by placing an ice cube in front of the flame. The ice cube melts and the hydrogen peroxide that forms during the reaction will dissolve in the water. You will collect water as it drips off the ice cube. Drippings will contain a small concentration of H2O2. The purpose of this experiment is for you to determine how much hydrogen peroxide is in it.
One way to calculate the amount of hydrogen peroxide present in a solution is to measure the amount of another chemical that reacts with it. This can be done using potassium permanganate (KMnO4), which forms a bright pink solution when dissolved in water. The permanganate ion and hydrogen peroxide are converted to Mn2+ and O2 gas, respectively, in a redox reaction. An acid is necessary in order to ensure that the reaction goes to completion.
As long as the melted ice solution contains hydrogen peroxide, the solution will not turn pink color as you add KMnO4. A slight pink color of the solution indicates that all the peroxide has reacted and now you are just adding excess KMnO4. If you know the amount of excess permanganate that you add, you can calculate the amount of MnO4- which reacted. Using the reaction stoichiometry, you can determine the original concentration of hydrogen peroxide in the drippings from the ice cube.
2MnO4(aq) + 5H2O2(aq) + 6H+(aq) --> 2Mn2(g) + 502(g) + 8H2O(l)
Mass of test tube and beaker: 117.93
Mass of test tube with melted ice: 124.84
Mass of melted ice: 6.91
Initial volume of buret: 18.8 mL
Volume of KMnO4 added.: 5.6mL
Volume of melted ice: 6.922 mL
Final Volume of buret: 13.2 mL
Total volume of solution: 15.6 mL
1. Calculate the Moles of excess MnO4- present. Show your work.
2. Calculate the moles of MnO4- which reacted.
3. Calculate the moles of H2O2 which reacted.
4. Calculate the moles of MnO4 added. Show work.
Explanation / Answer
2MnO4- (aq) + 5H2O2(aq) + 6H+(aq) --> 5O2(g) + 2Mn2+ (aq) + 8H2O(l)
Each 1 mole of KMnO4 provides 1 mole MnO4-, thus Molarity of MnO4- = Molarity of KMnO4
Molairy mnO4- = 0.105 M
Molarity = moles / litres
Therefore moles = molarity x litres
moles MnO4- = 0.105 M x 0.0432 L
= 0.004536 moles
= 0.00454 moles
The balanced equation tells you that
2 moles MnO4- reacts with 5 moles H2O2
So 1 mole MnO4- reacts with 5/2 moles h2O2
Thus 0.004536 moles MnO4- reacts with (5/2 x 0.004536) moles H2O2
= 0.01134 moles H2O2
= 0.0113 moles H2O2
moles = mass / molar mass
Therefore mass = molar mass x moles
mass H2O2 = 34.016 g/mol x 0.01134 moles
= 0.38574 g
= 0.386 g H2O2
mass percent = mass H2O2 / total mass x 100/1
= 0.386 g / 13.8 g x 100
= 2.80 % (3 sig figs)
Oxidation is loss of electrons. The species that undergoes oxidation loses electron. It is called the reducing agent because it causes the other reagent to be reduced
reduction is gain of electrons. The species that undergoes reduction gains electrons. It is called the oxidising agent because it takes the electrons from the other reagent thus causeing it to be oxidised.
Reduction half equation (note MnO4- is gaining electrons to Mn2+. It is reduced and thus is the oxidising agent)
2MnO4- + 16H+ + 10e ----------> 2Mn2+ + 8H20
oxidation half equation (note H2O2 is losing electrons thus being oxidised to O2. It is oxidised and thus is the reducing reagent)
5H2O2 -------> 5O2 + 10H+ + 10e
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.