Respuesta :
Answer : The moles of [tex]O_2[/tex] left in the products are 0.16 moles.
Explanation :
First we have to calculate the moles of [tex]CH_4[/tex].
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1 atm
V = volume of gas = 10 L
T = temperature of gas = [tex]27^oC=273+27=300K[/tex]
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:
[tex](1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)[/tex]
[tex]n=0.406mole[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
The balanced chemical reaction will be:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of [tex]CH_4[/tex] react with 2 moles of [tex]O_2[/tex]
So, 0.406 mole of [tex]CH_4[/tex] react with [tex]2\times 0.406=0.812[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the excess moles of [tex]O_2[/tex].
[tex]O_2[/tex] is 20 % excess. That means,
Excess moles of [tex]O_2[/tex] = [tex]\frac{(100 + 20)}{100}[/tex] × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × 0.812 = 0.97 mole
Now we have to calculate the moles of [tex]O_2[/tex] left in the products.
Moles of [tex]O_2[/tex] left in the products = Excess moles of [tex]O_2[/tex] - Required moles of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] left in the products = 0.97 - 0.812 = 0.16 mole
Therefore, the moles of [tex]O_2[/tex] left in the products are 0.16 moles.
