In comparing a 0.25 molality aqueous NaCl solution to a 0.25 molality aqueous CaCl 2 solution: a. the NaCl solution has the higher boiling point and the CaCl2 solution has the lower freezing point.
b. the NaCl solution has the higher boiling point and the lower freezing point.
c. the CaCl2 solution has the higher boiling point and the lower freezing point.
d. the CaCl2 solution has the higher boiling point and the NaCl solution has the lower freezing point.
e. both solutions have the same boiling point and the same freezing point.

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