- #1
Lori
Homework Statement
What mass of iron(3) chloride should be added to 450 ml of 0.132 M CaCl2 to give a solution with final chloride concentration of 0.555M
There are a lot of numbers, and I get confused on how to convert it to get mass of iron(3) chloride!
Homework Equations
M = n/L
The Attempt at a Solution
? g iron(3) chloride (162.2 g/mol)
.450 L of 0.132 M CaCl2 gives 0.0594 mols
0.555 M Cl- = 0.555 mols/L
I know the mole to mole ratio we would use is 2:3 but like, i get stuck at trying to start with the right numbers in my dimensional analyst...
my attempt:
0.555 mols/ L Cl- * (0.45 L CaCl2) * (3 mols Cl/2molsCl) *(1mol FeCl3/3molsCl ) = 0.12 (which is wrong).