chemistry question?

Question

4Al + 3O₂ → 2Al₂O₃
How does 4Al+3O₂ turn into 2Al₂O₃? WHat type of dividing is going on? Shouldn't it be 4Al3O₂? (the original problem was to make Al+O₂ into a balaced equation, and I understood it until that last part.) I know it has something to do with the fact that the charge of Al is +3 and the charge of O is -2 but I don't know what to do with that. Is +3 and -2 supposed to add up to something?Sorry I'm just really confused.

Answer 1

Hello Athena

 

Al₂O₃ is an ionic compound consisting of a metal, Al, and a nonmetal O.  In ionic compounds, metals will lose electrons to form positive ions (cations) while nonmetals will gain electrons to form negative ions (anions).

 

Since Al is in Group #13 (or Group 3A) of the main elements, it will lose 3 electrons to form Al³⁺ cation.

 

Since O is in Group #16 (or Group 6A) of the main elements, it will gain 2 electrons to form O^2- anion.

 

In a "neutral" ionic compound, the sum of the charges of the cations and anions MUST ALWAYS equal zero. 

Therefore, it you start with one Al³⁺ and one O^2-, the sum of the charges of these 2 ions is +3 -2 = +1.

 

If we have one Al³⁺ and two O^2- ions, the sum of the charges is +3 -2 -2 = -1

 

If we have two Al³⁺ and two O^2- ions, the sum of the charges is +3 +3 -2 -2 = +2

 

If we have two Al³⁺ and three O^2- ions, the sum of the charges is +3 +3 -2 -2 -2 = 0 !!

 

This means that the correct chemical formula for aluminum oxide should contain 2 Aluminum atoms and 3 Oxygen atoms, hence, the formula is Al₂O₃.

 

Admittedly, this is the longer process for writing chemical formulas, but the faster method is called the "cris cross" method.

 

To answer your question, let's first write the original reaction where Aluminum reacts with Oxygen to form aluminum oxide.

 

Al + O₂ → Al₂O₃

 

According to the Law of Conservation of Mass, the # of atoms of the elements on the left (reactant) side of the equation must equal the # of atoms on the right (product) side.  To "balance" the atoms, you have to use coefficients, which are whole numbers written in front of the chemical symbol for an element and in front of formulas for compounds.

 

Start by balancing O.  Left side has 2, right side has 3.  When you have odd-even numbers, the key is to find a common factor, which in this case is 6 (2 x 3). Place a coefficient of 3 in front of O₂ and a coefficient of 2 in front of Al₂O₃.

 

Al + 3O₂ → 2Al₂O₃

 

The oxygens are balanced, but Al is not.  Now let's balance Al.

 

Left side has 1, right side has 4 (2Al₂ is 2 x 2).  Place a 4 in front of the symbol for Al.

 

4Al + 3O₂ → 2Al₂O₃

 

The reaction is now balanced!  Even after completing this entire process, always double check to ensure the coefficients used are correct.

 

Hope the above was helpful!