Question

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is left in the products?

Answer 1

Answer : The moles of
`O_2` left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of
`CH_4`.

Using ideal gas equation:

`PV=nRT`

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas =
`27^oC=273+27=300K`

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

`(1atm)* (10L)=n* (0.0821L.atm/mol.K)* (300K)`

`n=0.406mole`

Now we have to calculate the moles of
`O_2`.

The balanced chemical reaction will be:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

From the balanced reaction we conclude that,

As, 1 mole of
`CH_4` react with 2 moles of
`O_2`

So, 0.406 mole of
`CH_4` react with
`2* 0.406=0.812` moles of
`O_2`

Now we have to calculate the excess moles of
`O_2`.

`O_2` is 20 % excess. That means,

Excess moles of
`O_2` =
`((100 + 20))/(100)` × Required moles of
`O_2`

Excess moles of
`O_2` = 1.2 × Required moles of
`O_2`

Excess moles of
`O_2` = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of
`O_2` left in the products.

Moles of
`O_2` left in the products = Excess moles of
`O_2` - Required moles of
`O_2`

Moles of
`O_2` left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of
`O_2` left in the products are 0.16 moles.