Question

Cl=8 F=22

(a) How many chlorine trifluoride molecules are formed in the reaction depicted above?

(b) How many chlorine molecules remain after the reaction has gone to completion?

(c) How many fluoride molecules remain after the reaction has gone to completion?

(d) Which is the limiting reactant in the reaction depicted above?

(e) If the reaction was carried out with 2.05 moles of chlorine gas and 4.46 moles of fluorine gas, how many moles of chlorine trifluoride would be produced?

(f) If the reaction was carried out with 2.05 g of chlorine gas and 4.46 g of fluorine gas, how many grams of chlorine trifluoride would be produced?

Answer 1

The formation of chlorine trifluoride (ClF₃) from chlorine molecule (Cl₂) and fluorine molecules (F₂) is

Cl₂ + 3F₂ ----------> 2ClF₃  

i.e 1 mole of  Cl₂ react with 3 moles of F₂ to from 2 moles of ClF₃  

Now given molecules of  Cl₂ = 8

So requited molecules of  F₂ from complete reaction = 8*3 = 24

But given molecules of  F₂ present = 22

That means we have less than required molecules of  F₂ . So here flourine is the limiting reagent. So the actual reaction wil be according to Flourine

i.e

mols of F₂ react = 22

mols of Cl₂ react = 22/3 = 7.33

mols of ClF₃ formed = 2/3 * 22 = 14.66

or 7.33 Cl₂ + 22 F₂ ----------> 14.66 ClF₃  

i.e 7.33 mole of  Cl₂ react with 22 moles of F₂ to from 14.66 moles of ClF₃  

a- So total 14.66 moles of ClF₃ will be formed

b- mols of Cl₂ left after complete reaction = 8 - 7.33 = 0.67‬ mols

c- mols of F₂ left after complete reaction = 0 mols

d- F₂ is the limiting reagenet

e-

If we take molecules of  Cl₂ = 2.05

So requited molecules of  F₂ from complete reaction = 2.05*3 = 6.15‬

But given molecules of  F₂ present = 4.46

That means we have less than required molecules of  F₂ . So here flourine is the limiting reagent. So the actual reaction wil be according to Flourine

i.e

mols of F₂ react = 4.46

mols of Cl₂ react = 4.46/3 = 1.486

mols of ClF₃ formed = 2/3 * 4.46 = 2.973

1.486 Cl₂ + 4.46 F₂ ----------> 2.973 ClF₃  

So finally 2.973 molecules of ClF₃ will be produced

f-

If we take mass of  Cl₂ = 2.05 g

Then mols of Cl₂ taken = mass / molar mass

= 2.05 g / 70 g/mol

= 0.029 mols

So requited molecules of  F₂ from complete reaction = 0.029 *3 = 0.087

But given mass of  F₂ present = 4.46 g

Then mols of F₂ taken = mass / molar mass

= 4.46 g / 38 g/mol

= 0.117 mols

That means we have more than required molecules of  F₂ . So here Chlorine is the limiting reagent. So the actual reaction wil be according to Chlorine

i.e

mols of Cl₂ react = 0.029 mols

mols of F₂ react = 0.029 mols * 3 = 0.087

mols of ClF₃ formed = 2 * 0.029 = 0.058‬ mols

So mass of ClF₃ formed = moles * molar mass of ClF₃

= 0.058‬ mols * 92.448 g/mol

= 5.36 g