How many L of CO2 at STP could be produced from the reaction of 2 L of CH4 at STP with 2 L of O2 at STP?

CH4+2O2-2H2O+CO2

A 1L

B 2L

C 3L

D 4L

Option A = 1 L

Explanation:

Given dа ta:

Volume of CH₄ = 2 L

Volume of O₂ = 2 L

Volume of CO₂ = ?

Solution:

First of all we will write balance chemical equation.

CH₄ + 2O₂ → 2H₂O + CO₂

Number of moles of CH₄:

PV = nRT

n = PV/RT

n = 1 atm. 2 L / 0.0821 atm. L. mol⁻¹. K⁻¹ * 273.15 k⁻¹

n = 2 atm. L / 22.43 atm. L. mol⁻¹

n = 0.09 mol

Number of moles of O₂:

PV = nRT

n = PV/RT

n = 1 atm. 2 L / 0.0821 atm. L. mol⁻¹. K⁻¹ * 273.15 k⁻¹

n = 2 atm. L / 22.43 atm. L. mol⁻¹

n = 0.09 mol

Now we compare the moles of O₂ and CH₄ with CO₂ from balance chemical equation

CH₄ : CO₂

1 : 1

0.09 : 0.09

O₂ : CO₂

2 : 1

0.09 : 1/2*0.09 = 0.045 mol

Volume of CO₂

PV = nRT

V = nRT / P

V = 0.045 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 273.15 k⁻¹ / 1 atm

V = 1.0092 L

Which is approximately equal to 1 L.