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6. A food supply company produces Yellow-42 (from previous example) in 200-gram wholesale bottles. How many grams of FeCl3 would be needed to produce one bottle of Yellow 42?


7. How much NaOH would be needed to produce one bottle of Yellow 42?


8. Iron (III) chloride is fairly expensive, and the company does not want to waste any during the reaction process. How could the company set up this reaction in a way that ensures no iron (III) chloride is wasted in the reaction?

Respuesta :

A food supply company produces Yellow-42 (from previous example) in 200-gram wholesale bottles. The grams of FeCl3 needed to produce one bottle of Yellow 42 is 303.638 grams.  The amount of NaOH needed to produce one bottle of Yellow 42 is 226.32 grams.

  • The mass amount of Yellow 42 = 200 grams
  • The molar mass of Fe(OH)3 needed to produce 1 bottle of Yellow 42 = 106.845 g/mol

Using the relation for the number of moles which is:

[tex]\mathbf{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

[tex]\mathbf{number \ of \ moles \ of \ Yellow \ 42 = \dfrac{200 g }{106.867 \ g/mol}}[/tex]

[tex]\mathbf{number \ of \ moles \ of \ Yellow \ 42 =1.872 \ moles}[/tex]

Now,

  • If 1 mole of Fe(OH)₃ requires 1 mole of FeCl₃
  • 1.872 moles of Fe(OH)₃ will require 1.872 moles of FeCl₃

Mass of FeCl₃ required = number of moles of  FeCl₃ × molar mass of FeCl₃

Mass of FeCl₃ required = 1.872 moles × 162.2 g/mol

Mass of FeCl₃ required = 303.638 grams

7.

Recall that:

  • 1 bottle of the yellow requires 1.872 moles of the Fe(OH)₃
  • Thus, 1 mole of Fe(OH)requires 3 moles of NaOH

Now, 1.872 moles of Fe(OH)will require (3 × 1.872 moles of NaOH)

= 5.616 moles of NaOH

Thus, the amount of NaOH in grams, that will be needed to produce one bottle of Yellow 42 is:

= 5.616 moles × 40.3 g/mol (molar mass of NaOH)

= 226.32 grams of NaOH

8.

The company could set up this reaction in a way that ensures no iron (III) chloride is wasted in the reaction by using the concept of the stoichiometry of the reaction involved in the system and he should feed the reactant within the ratio of 1:3 for FeCl₃ and NaOH respectively.

Learn more about stoichiometry here:

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