How is it possible for 10 grams of methane fuel to burn and emit 27 grams of carbon dioxide? write in paragraph or sentence form.
(the balanced equation is ch4 + 2o2 --> co2 + 2h2o)

             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.

CH4 +2 O2 → CO2 + 2 H2O

Explanation:

This reaction obeys the law of conservation of mass. The unbalanced equation CH4 + O2 → CO2 + H2O. 2 oxygen atoms on the left and 3 on the right side. There is 1 carbon atom on each side. 4 hydrogen atoms on one sides and 2 on the other. The Balanced equation is CH4 +2 O2 → CO2 + 2 H2O. In this equation the oxygen atoms balance with 4 oxygen atoms on one sides and 4 on the other. 1 carbon atom on each side. 4 hydrogen atoms on one side and 4 on the other side.

Therefore, every mole of methane combusted in excess oxygen will produce 1 mole of carbon dioxide. Since methane and carbon dioxide have molar masses of 16.04 g/mol and 44.01 g/mol, respectively, the mass expected from burning 10 g of methane in excess oxygen is:

10gCH4(1molCH416.04gCH4)(1molCO21molCH4)(44.01gCO21molCO2)=27gCO2

Therefore, yes, it is possible for 27 grams of carbon dioxide to produced from the burning of 10 g of methane fuel, though the actual yield is usually lower than the theoretical yield due to incomplete reactions, side reactions, etc.

Every balanced reaction equation must obey the law of conservation of mass. This occurs because there is no change to the nuclear structures of elements, only the arrangement of the elements. Therefore, each element has the same number of atoms as itself on either side of the reaction. Since the mass of an atom is consistent, the overall mass will not change, only the composition of the reactions.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

the dependant variable is spilled water

the independent is the heat