Question

How is it possible for 10 grams of methane fuel to burn and emit 27 grams of carbon dioxide? Please write in paragraph or sentence form.
(The balanced equation is CH4 + 2O2 --> CO2 + 2H2O)

Answer 1

Answer:

             First of all we will calculate the amount of CO₂ produced by burning 10 grams of Methane. The balance chemical equation is as follow:

                                CH₄  +  2 O₂   →    CO₂  +  2 H₂O

First we will calculate the moles of CH₄ as,

                            Moles  =  Mass / M.Mass

                            Moles  =  10 g / 16 g/mol

                            Moles  =  0.625 moles

Secondly calculate the moles of CO₂ produces theoretically,

According to equation,

                             1 mole of CH₄ produces  =  1 mole of CO₂

So,

               0.625 moles of CH₄ will produce  = X moles of CO₂

Solving for X,

                      X  =  1 mole × 0.625 moles / 1 mole

                     X =  0.625 moles of CO₂

At last calculate mass of CO₂ as,

                     Mass  =  Moles × M.Mass

                     Mass  =  0.625 mole × 44 g/mol

                     Mass  =  27.5 g ≈ 27 g

Conclusion:

                    From above calculation it can be concluded that this reaction is following law of conservation of mass. There is neither a gain nor a loss of mass during the reaction. Also, this can be made possible when the given amount of methane is provided with excess of O₂ gas so that the methane is completely combusted.