Part a - using the definition for mole fraction. a solution was prepared by dissolving 140.0 g of kcl in 305 g of water. calculate the mole fraction of kcl. (the formula weight of kcl is 74.6 g/mol. the formula weight of water is 18.0 g/mol.) express the mole fraction of kcl to two decimal places. view available hint(s) χ χ = nothing part b - use the definition of molality a solution was prepared by dissolving 140.0 g of kcl in 305 g of water. calculate the molality of kcl. (the formula weight of kcl is 74.6 g/mol. the formula weight of water is 18.0 g/mol.) express the concentration of kcl in molality to two decimal places. view available hint(s) express the concentration of in molality to two decimal places. 0.16 mol/kg 0.01 mol/kg 6.15 mol/kg 0.10 mol/kg

a) X KCl = 0.09

b) m KCl = 6.15 mol/Kg

Explanation:

a) mol fraction (X) = mol solute/mol total;  mol total = mol solute + mol solvent

⇒ X KCl = n KCl / n KCl + n H2O

∴ n KCl = 140.0g * ( mol / 74.6g ) = 1.876 mol KCl

∴ n H2O = 305g * ( mol / 18g ) = 16.944 mol H2O

⇒ X KCl = 1.876 / ( 1.876 + 16.944 ) = 0.09

b) molality (m) = mol solute / Kg solvent

∴ mol KCl = 140.0g KCl * ( mol / 74.6g ) = 1.876 mol KCl

∴ Kg H2O = 305g H2O * Kg/1000g = 0.305Kg H2O

⇒ m KCl = 1.876 / 0.305 = 6.15 mol/Kg

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