What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4 (g), reacts with excess oxygen? All gases are at 35.0C and 100.0 kPa.

V = 10.3 L

Explanation:

Given data:

Mass of methane = 6.40 g

Volume of CO₂ produced = ?

Temperature = 35°C (35+273 = 308 K)

Pressure = 100.0 KPa (100.0/101 = 0.98 atm)

Solution:

Chemical equation:

CH₄ +  2O₂        →       CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 6.40 g/ 16 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of CO₂ with CH₄.

         CH₄          :         CO₂

           1             :           1

        0.4            :         0.4

Volume of CO₂:

Formula:

PV = nRT

0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K ×  308 K

0.98 atm ×V = 10.11 atm.L

V = 10.11 atm.L /0.98 atm

V = 10.3 L

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