Fundamentals M15

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Nick Fiorentino 1E
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Joined: Wed Sep 18, 2019 12:16 am

Fundamentals M15

Postby Nick Fiorentino 1E » Thu Oct 03, 2019 3:07 pm

For problem M15, how would I go about this problem? Not really sure where to start

Jamie Lee 1H
Posts: 54
Joined: Fri Aug 09, 2019 12:15 am

Re: Fundamentals M15

Postby Jamie Lee 1H » Thu Oct 03, 2019 3:17 pm

You first need to start out by writing out a balanced equation from the elements that were given (aluminum metal + Chlorine gas --> aluminum chloride). From there, you would do the same calculations as the other limiting reactant problems to figure out whether chlorine or aluminum is a limiting reactant.

Posts: 99
Joined: Sat Sep 14, 2019 12:15 am

Re: Fundamentals M15

Postby VLi_2B » Thu Oct 03, 2019 3:23 pm

(a) First balance the chemical equation: Al + Cl2 -> AlCl3. It should look like this once it's balanced: 2Al + 3Cl2 -> 2AlCl3

(b)You need to find to the limiting reactant so you should 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. From the balanced equation, Al and Cl have a 2:3 ratio. The number of moles for Al is 225g/ 27g.mol^-1= 8.33mol and Cl2 has 535g/70.9g.mol^-1=7.55mol. You then observe that Cl2 is the limiting reactant in this case. Next, you can determine the mass of aluminum chloride using stoichiometry.
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

(c)Use the percent yield equation and plug in the numbers
Percentage yield= observed/theoretical x 100%
Percentage yield= 300g AlCl3/671g AlCl3 x 100%

Matt F
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

Re: Fundamentals M15

Postby Matt F » Fri Oct 04, 2019 10:11 am

^Thank you for this explanation, I found it very helpful. Will there be problems like this in the future where the molecules in the reaction are given by name as opposed to as the equation? (Ex: if they had given us Al(s) + Cl2(g) ---> AlCl3(s))

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