M.15

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donnanguyen1d
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M.15

Postby donnanguyen1d » Mon Oct 09, 2017 12:55 am

M.15 Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255 g of aluminum
is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300. g of aluminum chloride has been produced. (a) Write the balanced equation for the reaction. (b) What mass of aluminum chloride can be produced by
these reactants? (c) What is the percentage yield of aluminum chloride?

Chem_Mod
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Re: M.15

Postby Chem_Mod » Mon Oct 09, 2017 2:48 pm


this is the balanced equation...
Next, we must identify the limiting reactant. In order to do this you can solve for moles of Aluminum and moles of Chlorine and divide each one with their respective stoichiometric coefficients. The smaller number corresponds to the limiting reactant of the reaction. After working out the calculations, we find that Chlorine is the limiting reactant therefore we solve for the mass of aluminum chloride in the following way:

Next, we find percent yield which is defined as (actual yield/theoretical yield) x 100
Working this out for the percent yield of aluminum chloride we get:
percent yield = (300 grams/671 grams) x 100 = 44.7 %

Hope this helped Donna and I would love to answer any specific questions that you may have had about this problem. For future reference, it may be more helpful for you and other students to ask questions about specific concepts or topics you are having trouble with.

Jonathan Tangonan 1E
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Re: M.15

Postby Jonathan Tangonan 1E » Mon Oct 09, 2017 3:10 pm

(a) The chemical equation before being balanced would be Al + Cl2 -> AlCl3. After it is balanced it would look like 2Al + 3Cl2 -> 2AlCl3

(b)
For this one you have to determine the limiting reactant so you need to convert the masses of the reactants to moles and then compare the ratios of the stoichiometric coefficients to those of the number of moles present. Al and Cl have a 2:3 ratio while the number of moles for Al is 225g/ 27g.mol^-1= 8.33mol and Cl2 has 535g/70.9g.mol^-1=7.55mol. The 2:3 ratio in comparison to the 8.33mol : 7.55mol you can determine that Cl2 is the limiting reactant in this case.

After determining the limiting reactant you can then determine the amount of moles of AlCl3 produced by setting up the equation:

n of AlCl3 = 7.55mol of Cl2 x (2mol of Cl2/ 3 mol of AlCl3)
= 5.03 mol AlCl3
Mass of AlCl3= Mol of AlCl3 x Molar Mass of AlCl3
m= (5.03 mol)(133.33g.mol^-1)
= 670.64 g of AlCl3
=671g

(c)
From here you can determine the percentage yield by using the equation:

Percentage yield= observed yield/theoretical yield x 100%

Percentage yield= 300gAlCl3/671gAlCl3 x 100%
=44.7%

Yadira Flores 1G
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Joined: Wed Nov 15, 2017 3:01 am

Re: M.15

Postby Yadira Flores 1G » Thu Apr 05, 2018 4:29 pm

For the top answer in part b) when finding the limiting reagent I got 9.45mol for Al and 7.55mol for Cl2 but noticed that in the above posted answer they got 8.33mol. Did I make a calculation error?

NatalieSDis1A
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Re: M.15

Postby NatalieSDis1A » Sun Apr 08, 2018 1:21 pm

How do you come up with the equation Al + Cl2 --> AlCl3. I get that Cl is a diatomic molecule so it makes sense that we should use Cl2 but how do we know to put Cl3 in the product?

Chem_Mod
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Re: M.15

Postby Chem_Mod » Sun Apr 08, 2018 3:38 pm

You see that Al is a metal and Cl is a non-metal. So most likely AlCl3 is a ionic compound composed of aluminum ion Al3+ and Cl-. Thus, you need 3 Cl- to balance out Al3+ charge. If you are not sure on how to identify Cl as Cl- and Al as Al3+, you should probably read fundamental sections B-D also.


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