Double Displacement Reaction Calculations

Question

In our lab we completed conductometric titrations of FeCl3·6H2O + NaOH and CuCl2·2H2O + NaOH.

My question that has two parts:

1- If you could manipulate individual molecules and knew that you had 36 molecules of FeCl3, how many molecules of NaOH would you need to completely react with the FeCl3 and how many molecules of each of the products would you form?

2- What if you had 41 molecules of CuCI2?

Answer 1

FeCl₃(aq) + 3NaOH(aq) ==> Fe(OH)₃(s) + 3NaCl(aq) ... balanced equation

If you had 36 molecules (or moles or formula units, etc) it will take THREE TIMES as much NaOH according to the balanced equation or 108 molecules of NaOH. Using dimensional analysis, you can see this as follows:

36 molecules FeCl3 x 3 molecules NaOH/1 molecule FeCl3 = 108 molecules of NaOH

As for the products formed:

36 molecules FeCl3 x 1 molecule Fe(OH3/1 molecule FeCl3 = 36 molecules of Fe(OH)3 formed

36 molecules FeCl3 x 3 molecules NaCl/1 molecule FeCl3 = 108 molecules of NaCl formed

CuCl2(aq) + 2NaOH(aq) ==> Cu(OH)2(s) + 2NaCl(aq) ... balanced equation

36 molecules CuCl2 x 2 molecules NaOH/1 molecule CuCl2 = 72 molecules of NaOH neeeded

36 molecules CuCl2 x 1 molecule Cu(OH)2 / 1 molecule CuCl2 = 36 molecules Cu(OH)2 formed

36 molecules CuCl2 x 2 molecules NaCl / 1 molecule CuCl2 =72 molecules NaCl formed

You should be able to do the calculations using 41 molecules of CuCl2 by the same approach.