Olivia D.

asked • 03/07/16

If 354.75g of HCl react with excess of MnO2, how many miles of Cl2 form?

Can you help me with the equation to solve this?

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Beth R. answered • 03/07/16

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Because the question states that the reaction contains an excess of MnO2, we can assume all of the HCl will be used up in the reaction, so HCl will be the limiting factor.
 
1) The first step is to figure out the balanced equation for this reaction.
The reaction is: __ HCl + __ MnO2 ⇔ __ H2O +__ Cl2 + __MnCl2.  We need to add coefficients to each reactant and product so that the sides of the equation balance.  These numbers will give us the ratios in which molecules are consumed or produced.
2) Start with HCl: we have one mole of HCl on the reactant side, but 4 moles of Cl ( 2 moles of Cl2) on the product side, so we will put a 4 in front of HCl to balance the Cl on both sides.
4 HCl + __ MnO2 ⇔ __ H2O +__ Cl2 + __MnCl2
3) Next Look at H: We now have 4 moles of H on the reactant side but only 2 on the product side, so we will add a coefficient of 2 to the H2O on the product side to balance.
4 HCl + __ MnO2 ⇔2 H2O +__ Cl2 + __MnCl2
4) Now look at Mn: We have 1 mole of Mn on the reactant side, and 1 mole on the product side, so it is balanced.  O has 2 moles on the reactant side and 2 moles on the product side.
So... Our final balanced equation is:
4 HCl +  1 MnO2 ⇔2 H2O +1 Cl2 +1 MnCl2 (you don't need to write the coefficients in if they are =1, but I included them for clarity)
 
Now that we know the ratios, we need to determine the number of moles of HCl we are starting with.  We know we have 354.75g HCl...
5) Convert 354.75g HCl to moles.  To do this, we need to calculate the molecular mass of HCl: H=1.00794g/mol and Cl = 35.453 g/mol, so the molecular mass = 36.46094.
6) To convert grams to moles, we divide the number of grams of substance by the molecular weight, so in this case we have:
354.75g/36.46094 g/mol = 9.73 moles HCl.
7) Now that we have determined the number of moles of HCl we are starting with, we go back to the chemical equation we found in step 4 above, which tells us that for every 4 moles of HCl that goes into the reaction, 1 mole of Cl2 is produced, because HCl has a reaction coefficient of 4 while Cl2 has a coefficient of 1.
8) Based on the ratios above, we know that the number of moles of Cl2 produced is 1/4 the number of moles of HCl put in to the reaction, so moles Cl2=moles HCL/4 = 9.73/4 =2.43 moles
 
So... This reaction will produce 2.43 moles of Cl2
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Michael P.

Yup. I blew it. I was in a rush last night and forgot to check my results.
 
Thank you, Beth, noticing my error (and completing the calculation).
 
In thinking about my error and Beth's solution, I find that this problem is tricky.
 
If a student, you, do not know about or cannot find the equation that Beth presents in her first step,
 
HCl + MnO2 = MnCl2 + Cl2 + H2O
 
then you might not be able to get past square (step) one.
 
Given what we know from the statement of the problem, we only know HCl + MnO2 = Cl2.
 
We need more information.
 
This equation is not a simple neutralization of a metal oxide by an acid, such as 2HCl + CaO = CaCl2 + H2O. Neutralizations with HCl do not produce chlorine gas (Cl2). Notice that chlorine undergoes an oxidation: 2Cl- = Cl2 + 2e-. This is a redox reaction. So, the manganese, Mn+4 in MnO2, must undergo a reduction--a change in oxidation state, but to what state? Manganese has four other oxidation states. A good and correct guess is Mn+2.
 
HCl + MnO4 = Mn+2 + Cl2
 
I don't know what level of chemistry you have been studying, but if you have been studying redox reactions, you should be able to show that Mn+2 is the appropriate product.
 
You can combine the two reduction (half) reactions that involve MnO4-, permangenate, to produce a reaction for the reduction of Mn+4 to Mn+2, that is, MnO4 to Mn+2.
 
MnO2 + 4H+ + 2e- = Mn+2 + 2H2O   (+1.51v - 0.59v = +0.92v)
 
Combining the reduction of MnO2 with the oxidation of Cl- (-1.36v) gives us a sophisticated version of the reaction of interest:
 
MnO2 + 4H+ + 2Cl- = Mn+2 + Cl2 + 2H2O   (+0.92v - 1.36v = -0.44v)
 
This is one tricky part of the problem: Usually, an element does not appear as a product with two oxidation states (Cl- and Cl2), but there are chlorine ions in the acid solution that can be added both sides of the equation so that amounts of H+ and Cl- equivalent to HCl appear as a reactant.
 
MnO2 + 4H+ + 4Cl- = Mn+2 + Cl2 + 2H2O + 2Cl-
 
This form of the equation is equivalent to the equation that Beth presented and use in the further calculation.
 
The problem here is that there are two balance equations with two mole ratios. Do you use the number of moles of H+ in the HCl that enter into the reaction (4:1) or do you use the number of moles of Cl- that produce Cl2 (2:1)? Either one is correct if goal is to show that you know how to do stoichiometric calculations.
 
 
Another tricky part of the problem occurs if you guess that Mn+2 is the correct product and work to balance the resulting equation:
 
HCl + MnO2 = Mn+2 + Cl2
 
Manganese is balanced, so balance the chlorine:
 
2HCl + MnO2 = Mn+2 + Cl2
 
Balancing the chlorines is where I stopped paying attention. Doubling the Cl2 would not have caused a problem if I had checked the final result--the hydrogens and the charges.
 
Balance the oxygen:
 
2HCl + MnO2 = Mn+2 + Cl2 + 2H2O
 
Finally, balance the hydrogens and the positive charges:
 
2HCl + MnO2 + 2H+ = Mn+2 + Cl2 + 2H2O
 
Here we are again with a balanced equation with an alternative mole ratio (2:1), but part of the initial HCl does not enter the reaction to produce products (unless MnCl2 precipitates from the solution).
 
The balanced reaction tells us that, in solution, only half of the HCl is used to create the product Cl2. The other half of the HCl provides the Cl- in solution. So, we have 4.87 moles of HCl producing 2.44 moles of Cl2. Interestingly, it is the same answer.
 
One further interesting result about this problem: If you are familiar with redox reaction, you could combine the half reaction reduction potentials to calculate the net redox potential for this reaction (-0.44v) to find that, at equilibrium, very little of the products (Cl2) are formed. The low solubility of MnO2 further diminishes the amount of products formed.
 
 
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03/08/16

Michael P. answered • 03/07/16

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Olivia,
 
Start by writing out what you know: HCl + MnO2 = Cl2.
 
This equation is not balanced, so you have to do it for yourself.
 
Balance the elements. So, manganese, Mn, has to be on both sides. In its common ionic form, it has two positive charges: Mn+2. Note that this is different than its valence in manganese dioxide, MnO2.
 
HCl + MnO2 = Mn+2 + Cl2
 
Let's balance the oxygen on the left by adding water on the right. Note that I need two water molecules to balance MnO2.
 
HCl + MnO2 = Mn+2 + 2H2O + Cl2
 
Balance the chlorine next.
 
4HCl + MnO2 = Mn+2 + 2H2O + Cl2 
 
Finally, balance the chlorines and we are done.
 
4HCl + MnO2 = Mn+2 + 2H2O + 2Cl2
 
Now that you have the balanced equation, you can use the stoichiometry to calculate how many moles of Cl2 are formed.
 
Michael.
 
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