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Topic: I've been out of school for 20+ yrs so rusty....need help  (Read 5336 times)

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Offline Mike1970

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I've been out of school for 20+ yrs so rusty....need help
« on: March 23, 2015, 07:40:52 PM »
In a test tube a solution of:

KMnO4 + NaHSO3 ----> MnO2

I understand the K and Na are not important in the solution.
My problem is I am not grasping how to get from : MnO4 + HSO3 ---> MnO2

Can someone help and EXPALIN the process?
I know there are 2 half-reaction (Oxidation & Reduction) In this case:

Oxidation is from HSO3
Reduction is in Mn

Thanks

Offline billnotgatez

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #1 on: March 23, 2015, 08:04:32 PM »
KMnO4 + NaHSO3  :rarrow:  MnO2

Just to let you know that there are editing features in the posting window that allow better writing of equations.


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Offline billnotgatez

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #2 on: March 23, 2015, 08:45:40 PM »
@Mike1970
Do you think you could write a complete balanced equation?

Offline Mike1970

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #3 on: March 23, 2015, 08:53:29 PM »
Honestly....no.

I have the rules for balancing.
Balance O by adding H2O.
Balance H by adding H+.
and balance the charge by adding e-.

I'm just not getting it I guess.

Offline Corribus

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #4 on: March 24, 2015, 12:40:07 AM »
You can't do anything until you have a balanced chemical reaction. Balancing means the same amount of elements on each side of the  :rarrow: . The first thing you need to do is determine what chemical species are consumed and formed by the reaction. Don't let the K and Na throw you off. These are spectators and can be eliminated because these salts fully dissociate in aqueous solution. So what you really have is MnO4, HSO3, and MnO2. Some of those are ions, though, and so you need to determine what their charges are.

Importantly, there's another species you need to account for in the products, and you should be able to see what element it must include.

Let's start with that and work from there.

(By the way, you haven't said what the endpoint is here. Is this part of a homework question, and lab report, or something else? If the former two, it helps to have the exact wording of the question in its entirety.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Borek

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #5 on: March 24, 2015, 03:50:43 AM »
My problem is I am not grasping how to get from : MnO4 + HSO3 ---> MnO2

You have omitted charges here, and it is never a good idea, as balancing reaction equation means - between others - balancing charge. Of the three compounds listed two are ions, one is a neutral molecule.

Quote
Oxidation is from HSO3
Reduction is in Mn

This is unclear. Substance either is an oxidizing agent (and gets reduced during the reaction) or is a reducing agent (and gets oxidized during the reaction). What you wrote can be both read correctly and wrongly, so it is not clear if you got it right - or wrong.

So, what GETS oxidized during the reaction?

See if these help:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc
http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method
http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Mike1970

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #6 on: March 24, 2015, 06:49:18 AM »
I'll try to answer the questions.   This is for a lab. I will paste the procedures for this test tube below.


Add 9 mL of potassium permanganate (KMnO4) to test tube.
Add a solution of 0.1M sodium bisulfite (NaHSO3) to test tube until you observe a reaction. Process was to add drops 10 at a time until reactions was observed.  (Reaction was liquid turned white/milky with a solid brown/black solid at the bottom. That being MnO2. This is a Oxidation State of +4 according to a chart we were provided.

The questions are as follows (in this order for the assignment).

8. What was the unbalanced oxidation half-reaction in Test Tube 4?

A. SO32-:rarrow: SO42- + 4e-
B. SO32-:rarrow: SO42- + 3e-
C. SO32-:rarrow: SO42- + 2e-
D. SO32-:rarrow: SO42- + 1e-

9. What was the unbalanced reduction half-reaction in Test Tube 4?

A. MnO4- + 1e-   :rarrow: MnO2
B. MnO4- + 3e-   :rarrow: MnO2
C. MnO4- + 2e-   :rarrow: MnO2
D. MnO4- + 4e-   :rarrow: MnO2

10. What was the net ionic equation for the redox reactions in Test Tube #4?Referring to the table in the background of the lab manual may be helpful.

A. 2OH- + 2MnO4- + SO32-:rarrow: 2MnO42- + SO42- + H2O
B. 4H+ + 2MnO4- + 3SO32-:rarrow: 2MnO2 + 2H2O + 3SO42-
C. 2H+ + 2MnO4- + 3SO32-:rarrow: 2MnO2 + H2O + 3SO42-
D. 6H+ + 2MnO4- + 5HSO3-  :rarrow: 2Mn2+ + 5HSO4- + 3H2O

Offline Borek

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #7 on: March 24, 2015, 06:59:56 AM »
Have you tried to answer questions 8 and 9? They are well chosen small steps in obtaining the final equation.
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Offline Mike1970

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #8 on: March 26, 2015, 09:28:45 PM »
No, I have not. Reason being I did not know/understand how to.
But, I figured it out with the help of the High School teacher in my area. He taught me how to do them.
Not easy yet but at least I can make my way thru.

also found out that I was being tested on something that wasn't covered until the next chapter where it talks about it.  Always fun.


Offline Mike1970

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #9 on: March 26, 2015, 09:37:06 PM »
New question for you.

In the equation: 2Mg + O2   :rarrow: 2MgO
The number of moles of oxygen gas needed to react with 4.0 moles of Mg is _____?
A. 1.0 Mole
B. 2.0 Moles
C. 3.0 Moles
D. 4.0 Moles              My guess is 4.0 Moles
E. 6.0 Moles


How many grams of Magnesium are needed to react with 16g of O2?
A 0.50g
B. 1.0g
C. 12g
D. 24g                      Here my guess is 24g
E. 48g


So can you guys help me out and let me know if I'm right? If I'm not help me out with the right answer and how to figure it out.

Thanks




Offline Corribus

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #10 on: March 27, 2015, 12:07:23 AM »
If you get confused by the mole concept, realize that it is, at its most basic, just a way to conveniently measure amounts of atoms or molecules, which are usually very large numbers in every day practice. Just as you would not express or measure the distance between here and the moon in inches - although you could, if you so chose - you would not count or express the number of molecules in a glass of water. Rather, you would say the glass contains so many milliliters - or grams - of water. "Moles" is a way to connect the mass of a macroscopic substance to the number of individual microscopic elements in the substance.

Getting to your question: rather than asking how many moles of Mg reactive with molecules of oxygen, it may be easier, though equivalent, to ask how many atoms of magnesium reactive with every molecules of O2. The question does intentionally try to throw you off by using O2, each molecule of which contains two oxygen atoms. Don't get fooled. The discrete unit in question here is an oxygen molecule. According to the equation given, how many atoms of magnesium are required to react with each molecule of oxygen gas?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline curiouscat

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #11 on: March 27, 2015, 02:38:39 AM »
Sometimes I feel that the "mole" concept was unnecessarily mystified during my early Chemistry classes.

Even the definitions they used were so abstruse & opaque. The reference to C12 didn't really clarify things but obscured them.

It's nothing but a convenient name for 6.023E23 units of something.

If it's a kmol instead of a gmol it's only 1000 times more units of the same thing. Any thing really. 

Offline Borek

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #12 on: March 27, 2015, 03:48:48 AM »
Please start a new thread witha a new question.

Mole is just an overgrown dozen.

In the equation: 2Mg + O2   :rarrow: 2MgO
The number of moles of oxygen gas needed to react with 4.0 moles of Mg is _____?
A. 1.0 Mole
B. 2.0 Moles
C. 3.0 Moles
D. 4.0 Moles              My guess is 4.0 Moles
E. 6.0 Moles

Do you know how to read the reaction equation? Please read

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Mike1970

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Re: I've been out of school for 20+ yrs so rusty....need help
« Reply #13 on: March 27, 2015, 06:48:55 AM »
Please start a new thread witha a new question.

Mole is just an overgrown dozen.

In the equation: 2Mg + O2   :rarrow: 2MgO
The number of moles of oxygen gas needed to react with 4.0 moles of Mg is _____?
A. 1.0 Mole
B. 2.0 Moles
C. 3.0 Moles
D. 4.0 Moles              My guess is 4.0 Moles
E. 6.0 Moles

Do you know how to read the reaction equation? Please read

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations


Thank you for the link, That was better than the book I have.
I would change my answer to B.  2 moles.

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