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 correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol