# As a result of the combustion of ethylene C2H4, weighing 14 g, carbon (4) oxide was formed, weighing 44 g

**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;