Question

How many moles of chloride ion (Cl-) are present in a 395 mL of a 1.79 M solution of beryllium chloride (BeCl2)?

Answer 1

Answer: 1.414 mol Cl-  

Explanation:

First, let's find the moles of BeCl2.

1.79 M = 1.79 mol/L

395 mL = 0.395 L , since 1L = 1000 mL

Using stoichiometry,

1.79 mol/L *0.395 L = about 0.707 mol BeCl2

Now, we have 0.707 mol of BeCl2, but we are trying to find moles of Cl-

There are two chloride ions in each molecule of BeCl2, so there will be twice as many moles as well.

0.707 mol BeCl * (2 mol Cl- / 1 mol BeCl) = 1.414 mol Cl-  

Hope this helps!