Considering 0.10 m solutions of each substance, which contains the smallest concentration of ions

Question options:

A) K2SO4

B) FeCl₃

C) NaOH

D) NH₃

E) KCl

D. NH₃

Explanation:

K2SO4 = 2 K+ + SO42-

[K+]= 2 x 1.0 = 2.0 M ; [SO42-] = 1.0 M

total concentrations of ions = 2.0 + 1.0 = 3.0 M

FeCl3 = Fe3+ + 3Cl-

[Fe3+] = 1.0 M ; [Cl-] = 3 x 1.0 = 3.0

total concentration ions = 1.0 + 3.0 = 4.0 M

NaOH = Na+ + OH-

[Na+] = [OH-] = 1.0 M

total concentration ions = 1.0 + 1.0 = 2.0 M

NH3 is a weak acid so the concentration of NH4+ and OH- < 2.0

KCl = K+ + Cl-

[K+] = [Cl-] = 1.0 M

total concentration ions = 1.0 + 1.0 = 2.0 M

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g