Taking inyo account the stoichiometry of the reaction, when 189.6 g of ethylene burns in oxygen to give carbon dioxide and water, 594.9 grams of COâ‚‚ are formed.
First, taking into account that the Law of Conservation of Matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products, the balanced reaction is:
C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of each compound is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Then it is possible to apply the following rule of three: if by stoichiometry 28.05 grams of Câ‚‚Hâ‚„ produce 88.02 grams of COâ‚‚, 189.6 grams of Câ‚‚Hâ‚„ produces how much mass of COâ‚‚?
[tex]mass of CO_{2} =\frac{189.4 grams of C_{2} O_{4}x 88.02 grams of CO_{2}}{28.05grams of C_{2} O_{4}}[/tex]
mass of COâ‚‚= 594.9 grams
In summary, when 189.6 g of ethylene burns in oxygen to give carbon dioxide and water, 594.9 grams of COâ‚‚ are formed.
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