Question

17) Chlorine and Fluorine react to form gaseous chlorine trifluoride. You start with 1.75 mole of chlorine and 3.68 moles of fluorine. a. Write the balanced equation for the reaction. b. What is the limiting reactant? c. Find the moles of ClF3produced? d. Find the moles of excess left over

Answer 1

a.
`Cl_2+3F_2\rightarrow 2ClF_3`

b. fluorine

c. 2.45 moles

d. 0.52 moles

Step-by-step explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

a) The balanced chemical equation is:

`Cl_2+3F_2\rightarrow 2ClF_3`

Given : moles of chlorine = 1.75

moles of fluorine = 3.68

By Stoichiometry of the reaction:

3 moles of fluorine reacts with = 1 mole of chlorine

So, 3.68 moles of fluorine reacts with =
`(1)/(3)* 3.68=1.23moles` of chlorine

b) Thus fluorine is the limiting reagent as it limits the formation of product.

Chlorine is the excess reagent and (1.75-1.23) = 0.52 moles of chlorine are left over.

As 3 moles of fluorine give = 2 moles of chlorine trifluoride.

3.68 moles of fluorine will give =
`(2)/(3)* 3.68=2.45` moles of chlorine trifluoride.

c. Thus 2.45 moles of chlorine trifluoride
`ClF_3` are produced.

d. 0.52 moles of chlorine are left over.

Answer 2

a)
`Cl_2 + 3F_2 \rightarrow 2ClF_3`

b)
`F_2` is the limiting reagent

c) Moles of
`ClF_3` produced = 2.45 mol

b) Moles of
`Cl_2` left = 1.75 -1.22 = 0.53

Step-by-step explanation:

a) Balanced reaction:

`Cl_2 + 3F_2 \rightarrow 2ClF_3`

b)

No. of mole of
`Cl_2 = 1.75\ mol`

No. of mole of
`F_2 = 3.68\ mol`

`Cl_2 + 3F_2 \rightarrow 2ClF_3`

As, it is clear from the reaction that,

1 mol of
`Cl_2` requires 3 moles of
`F_2`

1.75 mol of
`Cl_2` require = 1.75 × 3 = 5.25 mol of
`F_2`

As, only 3.68 mol of
`F_2` is present, so
`F_2` is the limiting reagent.

c)

3 moles of
`F_2` form 2 moles of
`ClF_3`

3.68 moles of
`F_2` will form =
`(2)/(3)  * 3.68 = 2.45\ mol\ of\ ClF_3`

d)

`Cl_2` is present in excess.

3 moles of
`F_2` requires 1 mol of
`Cl_2`

3.68 moles of
`F_2` will require =
`(1)/(3) * 3.68 = 1.22\ mol\ of\ Cl_2`

`Cl_2` left = 1.75 -1.22 = 0.53 mol