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Report (any type)/Brief report
22/09/2021
The reactants of this particular combustion reaction are ##CH_4## and ##2O_2##.
Combustion reactions follow the general formula of ##”hydrocarbon”+”oxygen”=>”carbon dioxide”+”water”##.
Using the information we have, which is the amount of ##”carbon dioxide”##, or ##CO_2## and the amount of ##”water”##, or ##H_2O## we have to find the ##”hydrocarbon”## and the amount of ##”oxygen”## or ##O_2##.
must be balanced. The amount of ##C##s, ##O##s and ##H##s in this chemical equation must be equal on both the left and right sides. Our right side is complete, so we can simply count how many are on the right side so we can match it on the left.
On the right side we have. 1 carbon (##C##), 4 oxygens (##O##) and 4 hydrogens (##H##). Remember that the ##2## prefix in front of ##H_2O## means that both the hydrogen and oxygen in ##H_2O## doubles.
Therefore, using the ##C##s and ##H##s we can balance the left side with the right side by making a hydrocarbon. We have 1 carbon and 4 hydrogens, therefore our hydrocarbon is ##CH_4##. We typically call this methane.
Using the ##O##s on the right side we can figure out the ##O_2##s on the left side. We have ##4O##s on the right therefore we have ##2O_2##s on the left.
How did we know how to balance it like this? We counted up how many of each element was on the right side, and following the general formula of ##”hydrocarbon”+”oxygen”=>”carbon dioxide”+”water”## we used those amounts to balance with the above formula on the left side.
All in all, the reactants are ##CH_4## and ##2O_2## creating a full chemical equation of ##CH_4+2O_2=>CO_2+2H_2O##.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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