Consider the following balanced equation: Al(OH)3 + 3HCl mc005-1.jpg AlCl3 + 3H2O
Determine the number of grams of aluminum chloride, AlCl3, that are formed when 25.0 g of aluminum hydroxide reacts with 50.0 g of hydrochloric acid.
Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
This is a limiting reagent (LR) problem. You know that because more than one starting amount of material is given in the problem. First you must determine the LR.
mols Al(OH)3 = g/molar mass = 25/78 = 0.32
mols HCl = 50/36.5 = 1.37
How much AlCl3 will be formed just from the Al(OH)3. That's 0.32 moles.
How much AlCl3 will be formed just from HCl? That's 1.37/3 = 0.456 moles.
In LR problems the smallest amount will be the amount formed; therefore, Al(OH)3 is LR and 0.32 moles AlCl3 will be formed. Then
grams AlCl3 = moles AlCl3 x molar mass AlCl3 = ?