Answer : The moles of
left in the products are 0.16 moles.
Explanation :
First we have to calculate the moles of
.
Using ideal gas equation:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
where,
P = pressure of gas = 1 atm
V = volume of gas = 10 L
T = temperature of gas = ![27^oC=273+27=300K](https://tex.z-dn.net/?f=27%5EoC%3D273%2B27%3D300K)
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)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)](https://tex.z-dn.net/?f=%281atm%29%5Ctimes%20%2810L%29%3Dn%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28300K%29)
![n=0.406mole](https://tex.z-dn.net/?f=n%3D0.406mole)
Now we have to calculate the moles of
.
The balanced chemical reaction will be:
![CH_4+2O_2\rightarrow CO_2+2H_2O](https://tex.z-dn.net/?f=CH_4%2B2O_2%5Crightarrow%20CO_2%2B2H_2O)
From the balanced reaction we conclude that,
As, 1 mole of
react with 2 moles of ![O_2](https://tex.z-dn.net/?f=O_2)
So, 0.406 mole of
react with
moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Now we have to calculate the excess moles of
.
is 20 % excess. That means,
Excess moles of
=
× Required moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Excess moles of
= 1.2 × Required moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Excess moles of
= 1.2 × 0.812 = 0.97 mole
Now we have to calculate the moles of
left in the products.
Moles of
left in the products = Excess moles of
- Required moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Moles of
left in the products = 0.97 - 0.812 = 0.16 mole
Therefore, the moles of
left in the products are 0.16 moles.