Question

When solid phosphorus (P4) reacts with oxygen gas, diphosphorus pentoxide is formed. Initially 2.87 mol of phosphorus and 3.86 mol of oxygen are combined. (Assume 100% yield). a) Write a balanced equation for the reaction. b) What is the limiting reactant? c) How many miles of excess reactant remain after the reaction is complete?

Answer 1

(a) The balanced chemical reaction will be,

`P_4(s)+5O_2(g)\rightarrow 2P_2O_5(s)`

(b) The limiting reactant is
`O_2`.

(c) The remaining moles of excess reactant,
`P_4` is 0.94 moles.

Solution : Given,

Moles of
`P_4`= 2.87 mole

Moles of
`O_2` = 3.86 mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

`P_4(s)+5O_2(g)\rightarrow 2P_2O_5(s)`

From the balanced reaction we conclude that

As, 2 mole of
`O_2` react with 1 mole of
`P_4`

So, 3.86 moles of
`O_2` react with
`(3.86)/(2)=1.93` moles of
`P_4`

From this we conclude that,
`P_4` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the remaining moles of excess reactant,
`P_4`.

Remaining moles of excess reactant = Given moles of excess reactant - Required moles of excess reactant

Remaining moles of excess reactant = 2.87 - 1.93 = 0.94 moles

The remaining moles of excess reactant,
`P_4` is 0.94 moles.

Answer 2

a) The balanced equation for the reaction is as follows:

P4 + 5O2 -------> 2P2O5

b) The limiting reactant

The limiting agent has due to following properties:

I. It completely reacted in the reaction.

II. It determines the amount of the product in mole.

Given that

Phosphorus , P4 = 2.87 mole

Oxygen = 3.86 mole

Mole of P that

Phosphorus , P4 = 2.87 mole

Oxygen = 3.86 mole

Mole of P that

Phosphorus , P4 = 2.87 mole

Oxygen = 3.86 mole

Mole of P4 which reacted with the given mole of O are as follows:

3.86 mole O2 * 1 mole of P4 / 5 mole O2

=0.772 moles

Thus P4 is present in excess and O2 is limiting agent

c) the moles of excess reactant remain after the reaction is complete is P4 which is calculated as follows:

2.87 mole – 0.772 mol e= 2.098 mole