Quiz 1 Prep Fall 2015

[Josh_Zhong_1F]

Aluminum and hydrogen chloride, HCI, react to form aluminum chloride in solution, AlCl3, and hydrogen gas.
Balanced equation: 2Al + 6HCl -------> 2AlCl3 + 3H2
Question: If you had 5.43 g of aluminum and 7.80 g of hydrogen chloride, how many grams of AlCl3 would you get?


I could not figure out where to start and what to use.

Thanks!!

[Molly_McMillen_3J]

5.43g Al x (1 mole Al/26.982g Al) x (2 mole AlCl3/2 mole Al) x (133.341g AlCl3/1 mole AlCl3)= 26.8g AlCl3

7.8g HCl x (1 mole HCl/36.461g HCl) x (2 mole AlCl3/6 mole HCl) x (133.341g AlCl3/1 mole AlCl3)= 9.51g AlCl3

Since the final mass values are different depending on which reagent we base our calculations on, this means one of the reactants is a limiting reagent, in this case HCl is the limiting reagent because less AlCl3 is produced when calculating based on the mass of HCl we are given. Therefore, we pick the smaller mass of AlCl3, which is 9.51g, because that is the most that can be produced from what we are given. Hope this helps!

[Mirian_Garcia_2G]

Hello, so the first thing that should be done is balancing the chemical equation (which you have already done, yay!)
Next, you must take the grams of each element given and convert to moles using their molecular mass. You can then use your balanced chemical equation to develop a ratio that will give you the relative amount of moles of AlCl3 produced.

So, for Aluminum, the following will be the protocol:

(5.43 g Al)(1 mol Al/ 26.982 g Al)(2 mol AlCl3/2 mol Al)= 0.2012 mol AlCl3


This same process should be repeated with HCl in order to determine how many moles of AlCl3 can be produced. The one that produces less moles of AlCl3 is the limiting reactant. You will then convert the amount of moles of AlCl3 produced from the limiting reactant and acquire the number of grams of AlCl3 produced through using its molecular mass (133.341 grams/mol). This will be the grams of AlCl3 produced.

Hope I helped and wasn't too confusing!