What is the mass of nacl required to make 140 grams of a 12% solution of nacl in water?

a. 12 grams
b. 14 grams
c. 17 grams
d. 18 grams

C. 17 grams.

Explanation:

∵ mass % = [mass of solute/mass of solution] x 100.

mass of solute (NaCl) = ??? g & mass of solution = 140.0 g.

∴ mass of NaCl = (mass %)(mass of solution)/100 = (12.0)(140.0)/100 = 16.80 g ≅ 17.0 g.

mass of NaCl required = 17 grams

Explanation:

Mass% = mass of solute X 100 / total mass of solution

mass%= mass of solute X 100 / mass of solute + mass of solvent

given:

Mass of solution = 140 grams

mass % = 12%

Putting values

12 = mass of NaCl X 100 / 140

mass of NaCl = 12 X 140 /100 = 16.8 grams

mass of NaCl required = 17 grams

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