Question

When FeCl3 is ignited in an atmosphere of pure oxygen, the following reaction takes place- 4FeCl3(s) + 302(g) – 2Fe2O3(s) + 6C12(g) If 3 moles of FeCl3 are ignited in the presence of 2 moles of O2 gas, how much of which reagent is present in excess and therefore, remains unreacted?

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Answer 1

✿Answer :-

✿Given :-

✯When FeCl3 is ignited in an atmosphere of pure oxygen, the following reaction takes place :-

• 4FeCl3(s) + 302(g) – 2Fe2O3(s) + 6C12(g)

✯3 moles of FeCl3 are ignited in the presence of 2 moles of O2 gas.

✿Solution :-

In the Given reaction :-

3 Moles Of O2 gas Requires 4 Moles of FeCl3

✿ Thus O2 is the Limited reagent !!.

Thus 2 Mole of O2 will require :- 2×4/3 = 8/3= 2.67 Moles of FeCl3

Thus the Excess of the FeCl3 = 3-2.67 = 0.33 mol !!

✿ 0.33 mole FeCl3 remain Unreacted !!

Answer 2

Answer:

0.34 moles remains unreacted

Step by Step explanation :

From equation :-  

4 moles of $FeCl_{3}$ needs 3 moles of $O_{2}$

3 moles $FeCl_{3}$ will need =  $\frac{3 X 3}{4}$

                                      = 2.25 moles

But only 2 moles of $O_{2}$ is given

∴ $O_{2}$ is in limited quantity

$4FeCl_{3}$ = $3O_{2}$

$xFeCl_{3}$ = $2O_{2}$

2 X 4 = 3x

x = 2.66 moles

$FeCl_{3}$ remaining is = 3 - 2.66

                               =  0.34 moles unreacted

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