Calculate the number of moles of Cl− ions in 3.00 L of 0.230 M AlCl3?

Answer 1

2.07 moles is the correct answer.

Answer 2

Moles AlCl3 = liters X Molarity = 3.00 l X 0.230 moles / liter = 0.690 moles

AlCl3(s) ---------> Al+++(aq) + 3 Cl-(aq)

So the moles of chloride ion is three times the initial molarity of the aluminum chloride.

Moles of Cl- = 3 X 0.690 moles = 2.07 moles

Answer: 2.07 moles of Cl-

Answer 3

Volume (L) x concentration AlCl3 (mol/L) x (3 mol Cl-)/(1 mol AlCl3)

In words, volume times concentration gives you amount, but you also need to remember that each AlCl3 gives you three chlorides.