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 data:

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.

is there any options? maybe im just not seeing them but i need the options to

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