Question

In our lab we completed conductometric titrations of FeCl3 + 6H2O and CuCl2+2H2O causing each of them to undergo double displacement reactions.

I have a question that has two part:

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?

thanks! showing work would be extremely helpful

Answer 1

1) The question can be answered on the basis of the balanced stoichiometric equation.

FeCl₃ (aq) + 3 NaOH (aq) --------> Fe(OH)₃ (s) + 3 NaCl (aq)

Stoichiometric equations are written in terms of number of moles; however, conditions remaining same, we can approximated stoichiometric equations in terms of number of molecules.

1 mole of FeCl₃ requires 3 molecules of NaOH; therefore, 36 molecules of FeCl₃ will require (3*36) = 108 molecules of NaOH.

The reaction will produce 36 molecules of Fe(OH)₃ and 108 molecules of NaCl (ans).

2) The question can be answered on the basis of the balanced stoichiometric equation.

CuCl₂ (aq) + 2 NaOH (aq) --------> Cu(OH)₂ (s) + 2 NaCl (aq)

Stoichiometric equations are written in terms of number of moles; however, conditions remaining same, we can approximated stoichiometric equations in terms of number of molecules.

1 mole of CuCl₂ requires 2 molecules of NaOH; therefore, 41 molecules of CuCl₂ will require (2*41) = 82 molecules of NaOH.

The reaction will produce 41 molecules of Cu(OH)₂ and 82 molecules of NaCl (ans).