Question

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

Answer 1

A. fluorine, 1.79 moles

Step-by-step 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 =
`(mass)/(molar mass)`

C; molar mass = 12;

Number of moles =
`(87.7)/(12)` = 7.31moles

F; molar mass = 2(19) = 38g/mol

Number of moles =
`(136)/(38)` = 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
`(3.58)/(2)` = 1.79moles of CF₄