A sample of 87.6 g of carbon is reacted with 136 g of fluorine gas to produce carbon tetrafluoride. Using
the balanced equation below, predict which is the
limiting reactant and the maximum amount in moles
of carbon tetrafluoride that can be produced.
C +2F2 → CF4
A. fluorine, 1.79 moles
B. carbon, 7.29 moles
C. fluorine, 6.72 moles
D. carbon, 4.63 moles

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon  = 87.7g

Mass of fluorine gas  = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced  = ?

Solution:

   Equation of the reaction:

             C    +   2F₂ →   CF₄  

let us find the number of the moles the given species;

  Number of moles =  

  C;   molar mass = 12;

            Number of moles  =   = 7.31moles

 F;  molar mass  = 2(19)  = 38g/mol

             Number of moles  =   = 3.58moles

 So;

   From the give reaction:

          1 mole of C requires 2 moles of F₂

         7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

  Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

 So;

       2 moles of F₂ will produce  mole of CF₄  

       3.58 moles of F₂ will then produce  = 1.79moles of CF₄

the awnser is group 12

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