Write a balanced equation for the combustion of gaseous ethylene (C2H4), an important natural plant hormone, in which it combines with gaseous oxygen to form gaseous carbon dioxide and gaseous water
this answer is only partially correct... the correct answer is
C2H4(g)+3O2(g)-->2CO2(g)+2H2O(g)
C2H4 + O2 ---> CO2 + H2O
Balancing, it becomes
C2H4 + 3O2 ---> 2CO2 + 2H2O
How many grams of carbon dioxide were formed
To write a balanced equation for the combustion of gaseous ethylene (C2H4) with gaseous oxygen (O2) to form gaseous carbon dioxide (CO2) and gaseous water (H2O), we need to ensure that the number of atoms of each element is the same on both sides of the equation.
1. Begin by identifying the elements present in the reactants and products:
Reactants: C (carbon), H (hydrogen), O (oxygen)
Products: C (carbon), H (hydrogen), O (oxygen)
2. Count the number of atoms for each element on both sides of the equation and compare them:
Reactants: 2 carbon atoms, 4 hydrogen atoms, 2 oxygen atoms
Products: 1 carbon atom, 2 hydrogen atoms, 3 oxygen atoms
3. Balance the carbon atoms:
Since there are two carbon atoms in the reactant (C2H4) and only one in the product (CO2), we need to balance the carbon atoms on the product side. To do this, we put a coefficient of 2 in front of the carbon dioxide (CO2) on the product side.
C2H4 + O2 -> 2CO2 + H2O
4. Balance the hydrogen atoms:
Since there are four hydrogen atoms in the reactant (C2H4) and only two in the products (H2O), the hydrogen atoms are already balanced.
C2H4 + O2 -> 2CO2 + H2O
5. Balance the oxygen atoms:
Since there are two oxygen atoms in the reactant (O2) and three in the products (2CO2 + H2O), we need to balance the oxygen atoms on the reactant side. To do this, we put a coefficient of 3/2 in front of the oxygen (O2) on the reactant side.
C2H4 + (3/2)O2 -> 2CO2 + H2O
Therefore, the balanced equation for the combustion of gaseous ethylene (C2H4) with gaseous oxygen (O2) to form gaseous carbon dioxide (CO2) and gaseous water (H2O) is:
C2H4 + (3/2)O2 -> 2CO2 + H2O