If 24.9 moles of hydrochloric acid are reacted with excess aluminum, how many moles of aluminum chloride could be formed? (SHOW WORK)

Question

60. When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced.

2Al(s) + 6HCl(aq)  2AlCl3(aq) +3H2 (g)

If 24.9 moles of hydrochloric acid are reacted with excess aluminum, how many moles of aluminum chloride could be formed? (SHOW WORK)

Answer 1

Step 1: Usually you will need to determine what the limiting reactant is. HCl is going to be your limiting reactant in this problem since the aluminum is being added in excess.

Step 2: If you look at your balanced equation, using the stoichiometric coefficients of each molecule, you'll notice that it takes 6 moles of HCl to produce 2 moles of aluminum chloride. Using this information, we can write a conversion factor to do our stoichiometric calculations. If you're struggling with writing conversion factors, use this simple formula to write them:

Given Unit x (Desired Unit / Given Unit) = Desired Unit

Notice that when you take your given unit (in this case its moles of HCl) and divide it by the stoichiometric coefficient of HCl (6), the "given units" will cancel and leave you with your desired unit (moles of aluminum chloride). The whole idea behind stoichiometry is to reflect the quantity of one substance as a quantity of another substance using these coefficients. In this case, you want to know how much aluminum chloride can be made from a given amount of HCl. So you would set up your calculation like so:

24.9 moles HCl (2 moles AlCl₃ / 6 moles HCl) = 8.3 moles of AlCl₃ are produced

(you multiply the 24.9 moles of HCl by 2, and then divide the resulting number by 6 to get your answer)

If you wanted to take this a step further and calculate the mass of AlCl₃ produced, you would use the molar mass of aluminum chloride and convert from moles to grams: (molar mass is expressed in grams/mole, so you want the mole units to cancel so you are left with grams).

8.3 moles AlCl₃ (133.34 grams / 1 mole of AlCl₃) = 1,106.72 grams = 1.10672 kg