As a result of the combustion of ethylene C2H4, weighing 14 g, carbon (4) oxide was formed, weighing 44 g and water, weighing 18 g. Make an equation for this reaction and calculate the mass of oxygen that was spent.
The oxidation reaction of ethylene with oxygen is described by the following chemical reaction equation:
C2H4 + 3O2 = 2CO2 + 2H2O;
3 moles of oxygen are reacted with 1 mole of ethylene. This synthesizes 2 mol of carbon monoxide.
Let’s calculate the available chemical amount of ethylene substance.
M C2H4 = 12 x 2 + 4 = 28 grams / mol; N C2H4 = 14/28 = 0.5 mol;
Let’s calculate the amount of carbon dioxide substance.
M CO2 = 12 + 16 x 2 = 44 grams / mol; N CO 2 = 44/44 = 1 mol;
This reaction will require 0.5 x 3 = 1.5 mol of oxygen.
Let’s calculate the weight of oxygen.
M O2 = 16 x 2 = 32 grams / mol; m O2 = 32 x 1.5 = 48 grams;
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