Question

When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol of FeC13 are ignited in the presence of 2.00 mol of 0 2 gas, how much of which reagent is present in excess and therefore remains unreacted

Answer 1

Answer : The reagent present in excess and remains unreacted is, $O_2$

Solution : Given,

Moles of $FeCl_3$ = 3.00 mole

Moles of $O_2$ = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)$

From the balanced reaction we conclude that

As, 2 moles of $FeCl_3$ react with 1 mole of $O_2$

So, 3.00 moles of $FeCl_3$ react with $\frac{3.00}{2}=1.5$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $FeCl_3$ is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, $O_2$