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 CO₂ 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 C₂H₄ produce 88.02 grams of CO₂, 189.6 grams of C₂H₄ produces how much mass of CO₂?
[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 CO₂= 594.9 grams
In summary, when 189.6 g of ethylene burns in oxygen to give carbon dioxide and water, 594.9 grams of CO₂ are formed.
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