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### Can't Find Limiting Reactant

Posted: Thu Oct 10, 2019 11:05 pm
For the following reaction, how many grams of AlCl3 would be obtained if 5.43g of aluminum and 7.80g of hydrogen chloride (HCl) was used in the reaction?
Al (s) + HCl (aq) → AlCl3 (aq) + H2 (g)

I keep getting the calculations wrong? can someone explain step by step?

### Re: Can't Find Limiting Reactant

Posted: Thu Oct 10, 2019 11:29 pm
Hello, you would begin by balancing. Then you would use the two molar masses to convert both of the reagents to moles and continue to convert that based on the stoichiometric coefficients and molar mass of the product to see which creates less product. This is not too in detail, but I hope it helps and if necessary I can expand!

### Re: Can't Find Limiting Reactant

Posted: Thu Oct 10, 2019 11:31 pm
hey!
You first have to balance the equation = 2AL + 6HCL --> 2ALCL3 + 3H2.
You then calculate how many grams of product does each reactant give by calculating molar masses and using molecular weights of each.
Using Al, we get:
(5.43gAl/26.981g/molAl)*(2mol/2mol)*133.481g/molAlCl3 = 26.86g of AlCl3
Doing the same with HCL,
we get 9.506g. Since HCL produces lesser product, it is the limiting reactant.

### Re: Can't Find Limiting Reactant

Posted: Fri Oct 11, 2019 12:06 am
Make sure to balance first
It is a common mistake people make and a lot of teachers put it in like that in test in order to trick you.

### Re: Can't Find Limiting Reactant

Posted: Sun Oct 13, 2019 6:56 pm
You would first balance the equations and then convert the grams of each compound into moles. Then you would see which reactant would yield less of the product. By doing this you will see which reactant is limiting the mass of the product.