Chemistry question that I did but my tutor showed me and I dont think he showed me the easiest way and he?

Question

confused me because he does not know stoichmetry.

given reaction

2KMnO4(aq) + 15 HCl(aq) ----> 2MnCl2(aq) + 5Cl2(g) + 2KCl(aq)

1) calculate the moles of MnO4 ^-1 ions that are present in a .500 P 0f 2.05 M KMnO4 solution

okay here is what my "tutor" showed me

(2.05 m KMnO4/1L)(.500L)= 1.025 m KMnO4

0.500 L x (2.05 m KMnO4/1L)(1 m MnO4/1m KMnO4) = .025 m MnO4^-1

2) calculate the volume in mL of 6.00 M HCl required to react with 200. mL of .550 M KMnO4

(this is the part that confused me by what he did?? can you show an easier way?) im also confued by the m's? are they big or little? M or mol ?

200mL KMnO4 x (1.0550 m KMnO4/1000mL KMnO4)(16 m HCl/2 m KMnO4) = .88 m? HCl

(6 mol HCl/1000 mL) x (2 m KMnO4/16 m HCl) = .75 m KMNO4

(.75)(V1)=(.550 mL)(200 mL) = 146.6 mL

?

(.88)(1000mL)=(6)V2

I have no idea what is going on in all that ... like I said he goes over my head

heres the next part of the problem

2) calculate Cl2 moles produced from part b

Answer 1

I'm a little confused by your questions but here we go:

Molarity's (M) definition is  M= # mol/L and can be represented as m/L or M. When finding moles or L or M you can:

(This can be calculated even if 1L is not used:)

250 mL and 5 moles= 5 mol/ .250 L = 20 M

or

5 M solution= ? moles/ .10 L     moles= .5

The way you use it in stoichiometry, if you prefer to write it into a longer equation would be:

Element X's M is: 2 mol/L

1.50 moles X x (1 L / 2 moles X) =.75 L

There really isn't an easier way. You have to understand that M represents mol/L.

As for your C1V1=C2V2 equation, it's in reference to diluting a solution therefore, in order to calculate the new volume you take the initial volume and molarity and multiple together and then divide by the concentration on the other side to get the volume.

Think of it as a proportion which must isolate for the unknown variable. You'd understand if it was:

2 x 4 = 8 x ?

? = 1 because 2x4 = 8 then divided by the 8 on the other side = 1

As for the partial pressure...you must use the mole fraction to calculate this...

however...I don't know why it would be listed for water, so I cannot help :( It seems to have nothing to do with the equation of the reaction, other than the Cl2

Answer 2

IMPORTANT! First of all, your "balanced equation" is messed up! There are only 15 HCl's on the reactant side, but there are 16 Cl's on the product side, also,what happened to all the hydrogens from the HCl's and the oxygens from the MnO4's (probably formed water).

I think your equation should be :

2 KMnO4 + 16 HCl --> 2MnCl2 + 5 Cl2 + 2 KCl + 8 H2O

Please double check that this is the reaction you are working with, I am having to guess.

For the first one (1):

All you need is volume of the solution (in liters) x the molarity (which you already have) so as you have, there is 1.025 mole of KMnO4. Since there is one K+ and one MnO4- per KMnO4, the moles of MnO4- are the same as the moles of KMnO4.

Actually you tutor did it right, (the stuff in the second set of parenthesis is the one mole of MnO4- per one mole of KMnO4 part which I described in words) you just forgot the "1" before the .025. Because this is a simple example, a lot of the numbers are '"1's" and you can do it in your head, but it's good to practice the full calculation, so you know how it works for when you get more difficult questions.

For the second part (2)

Note: one place you say conc is 0.550M and another say it is 1.0550M for the KMnO4 and in one of the equations use 16-2 for HCl to KMnO4 ratio (doesn't match your equation)!!

First, you need to determine how many moles of HCl are required to react with the specified quantity of KMnO4. So we need moles of KMnO4:

0.200L x 0.550M = 0.110moles KMnO4

Now, we need the ratio from the balanced equation: 16 HCl's per 2 KMnO4

0.110moles x (16 HCl/2KMnO4) = 0.880moles HCl required.

Now, we need to find what volume of 6.0M HCl contains 0.880 moles:

0.880mole/6.0 mole/L= 0.147L or 147ml

I have to go now, huge thunderstorm!!!

Will try to get back tomorrow to help!!

OK, I'm back (wild thunderstorm!!, our 7th one for the day; I was afraid they would find my calcined bones next to a vaporized computer)

To get the volume of Cl2 produced (presumably from the 147 ml of HCl used in the last part,although you didn't specify, there doesn't seem to be a part "b" only 1 & 2) you just use moles of reactant and ratio from the balanced equation to get moles of product:

0.880 mole HCl x 5Cl2/15HCl = 0.293 mole Cl2

Because the ratio lists Cl2 not Cl on the top your answer will be in moles of Cl2 not Cl.

(REMEMBER! Double-check to see that the equation I provided is the correct one so you get the right answers.

The general strategy to find these answers is the same. Determine moles of starting reactant(s), determine if there is a limiting reagent, find ratio from BALANCED equation, calculate answer in moles and finally, convert moles to what ever other unit necessary to keep your chemistry teacher happy.

I hope this is of some use to you, I have to go now, we were half way through putting a new roof on when the storms hit!

For your last part: the total pressure of the gas mixture is equal to the sum of the partial pressures so, if the pressure of the Cl2 + the water vapor = 758mmHg and the water vapor at 16 C is 14mmHg then the partial pressure of the Cl2 is: (758 - 14)mmHg = 744mmHg

To get moles of Cl2 from this data: Note: determining moles of Cl2 from the data further back in the problem seems to give a different answer.

PV = nRT; solve for n:

n = PV/RT

P= 744mmHg/660mmHg/atm = 0.979atm

T = (16 + 273) = 289 K

n = 0.979atm x 0.425L/0.08208 Latm/mol K x 289 K = 0.0175 mole

Answer 3

Where to start? Hmmm.

First of all, when communicating with other people, you need to use what language you and they have in common.  In this case it is the "language" of chemistry.  When in doubt, take a look at the IUPAC conventions for terminology.  They may have something for a mole other than mole, but if so I am unfamiliar with it.  m means milli or 10^(-3) it is a standard SI prefix, I'm surprized you are unaware of that.  M is the only abbreviation I am aware of for molar (moles per liter) and l (lower case L) is the correct abbreviation for liter - however since l and 1 can be confused, (not to mention an upper case i ( I ), it is bettter to be clear than correct and so using mL for milliliter is good.  C is the abbreviation for centigrade (or is it Celsius?), not c.  OK, now that I've covered stuff I'd be surprised that you haven't covered we can move on to wtf a 0.500 P solution is and how 15 HCl on one side and multiples of 2 Cl on the other can be said to balance.  I "think" 15 is an odd number.  I see that your tutor is using 16 to solve the problem.  Perhaps its that 0.500 P solution ?

And then we have 0.55 or is it 1.055 or some other c**p?

So lets balance the equation, shall we?MnO4(-1) the Mn has a formal Ox St of +7 (= -(-2*4 -(-1)) and in MnCl2 it is +2 so each Mn gained 5 electrons.  And each Cl in HCl going to CL2 losses 1 electron - - So since we need 2 Cl(-1) ions for each Mn and 5 Cl(-1) ions to provide the electrons for each AND one Cl for the K.  Thats 8 Cl for each Mn.  With 2 KMnO4 we need 16 CL (and yes I should be typing Cl, but my fingers are old and slow).  Finally, we've neglected the H and O, which form H2O (8H2O to be precise)

so the mole ratio for the respective materials are

2: 16::2:5:2:8

And half a liter of 2.05 moles per liter gives us 1.025 moles

And one fifth of a liter of 0.55 M gives us 0.11 moles (part b ????????? sloppy again) and the respective ratio is 2:16

or 1:8 or 0.11 to 0.88 moles and to get 0.88 moles starting with 6 M we'd need --oh I bet you can do the math! (hint

moles divided by moles per liter gives units of liters)

and finally 0.11 at a ratio of 2:5 means we need .275 moles of Cl2 for it to balance, right? (2 is to 5 as 0.11 is to ...?)

Answer 4

In the second question about collected gas, if you collect a gas that has bubbled through and is over water, the gas is saturated with water vapor which has a vapor pressure related to temperature.   You have to subtract this pressure from the measured pressure so you can use the measured volume in calcs.  

In the first problem, I don't know what you mean by a "P"

In the second part, I think the tutor is "cramming" too much into one math line.  In the problem, "M" is the abbreviation for molarity, which means moles of solute per liter of solution.  "mL" is milliliter.  

It looks like he uses "m" for moles.   He likes to show units, which is good.  The calc should go:

(find moles of KMnO4 availiable):

= volume x molarity = 0.2 L x 0.55 mol/L = 0.11 moles.

(find moles of HCl needed to react).  The rxn tells us that the moles of HCl and KMnO4 need to be in the ratio of 15:2.  Then the moles of HCl in this case are  moles HCl/0.11 moles = 15/2=0.825 moles HCl

(find volume of HCl):  Since we use 6 M HCl, then

moles = volume x molarity, and

0.825 m= 6V,  V= 0.138 mL

Answer 5

m should be M which is shorthand for mol/L (it is a concentration)

when you're first learning this it sometimes helps to write out a table so you can see the steps of the calculation you're going to do, like so:

------------------------------------ 2KMnO4(aq) + 15 HCl(aq) ---->

volume (mL)  ------------------|  200              |    v(HCl)

concentration (M = mol/L)---- |  0.550           |    6.00

amount (mol) -----------------|   n(KMnO4)   |  n(HCl)

the only equation you need here is n = c*v so if you have two of those values you can work out the third one. so it should be clear that the first thing to do is to work out the amount of KMnO4, n(KMnO4):

n(KMnO4) = (0.550 mol/L) * (200 mL) / (1000 mL/L)

= 0.11 mol

now that reacts with HCl in the ratio 2:15 so that

n(KMnO4)/2 = n(HCl)/15

hopefully you can see that this equation works for any amounts, for example if you had 2 moles of KMnO4 and 15 moles of HCl the equation above would reduce to 1 mol = 1 mol

rearranging in terms of the amount of HCl, n(HCl)

n(HCl) = n(KMnO4) * 15/2 = 0.11 * 15/2 = 0.825 mol

now you have the amount and concentration (n and c) of HCl so you can work out the volume as required:

v(HCl) = n(HCl) / c(HCl) = (0.825 mol) / (6 mol/L) * (1000 mL / L) = 137.5 mL

I'm really not sure if that's what your tutor gets, it's not clear to me what his answer is. hopefully you can follow that better. there are faster ways to do it, but it's best to be able to follow what you're doing.

b) from stoichiometry:

n(KMnO4) / 2 = n(Cl2) / 5

rearranging:

n(Cl2) = n(KMnO4) *5/2 = 0.11 mol * 5/2 = 0.275 mol.

that last bit is right. actually the only information you need to use there is the total (air) pressure and the water vapour pressure. the setup in that type of experiment is called a gas burette. you have a sealed flask where the reaction takes place, connected to a closed tube, which is again connected to another tube which is open to the air. the two tubes are half filled with water so that the reaction flask is sealed by the water in the tubes. you keep the water levels in the tubes equal, so that the total pressure of the gas in the sealed compartment is the same as the air pressure (no added pressure due to displaced water). as gas is evolved by the reaction you can measure the volume change in the tubes. anyway the key point here is that the volume added to the sealed compartment is in contact with water so it will absorb water vapour, so that the chlorine gas which occupies that volume is at a slightly lower partial pressure than it would otherwise be: 758 mm Hg - 14 mm Hg = 744 mm Hg. of course the chlorine gas will diffuse so that it fills the entire volume of the sealed container, which also contained air, but that part of the sealed flask can be ignored for the purposes of the calculation. it does take a little thought to appreciate why this works...