What is the molarity and molality of a solution that is 10.00 % by mass potassium hydrogen carbonate (KHCO3, 100.11 g/mol) and has a density of 1.0650 g/mL

The molarity and molality of a solution is 1.06 M and 1.11 m respectively.

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

where,

n = moles of solute  =

= weight of solvent in g

10.00 g of potassium hydrogen carbonate in 100 g of solution

mass of solvent = mass of solution - mass of solute = (100.0-10.00) = 90.0 g

2. Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,

n= moles of solute  = 0.09985

= volume of solution in ml

Density of solution = 1036.5 g/L

Volume of solution =

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