Question Video: A Balanced Equation for Ammonia Oxidation | Nagwa Question Video: A Balanced Equation for Ammonia Oxidation | Nagwa

Question Video: A Balanced Equation for Ammonia Oxidation Chemistry • First Year of Secondary School

Reaction of ammonia (NH₃) with oxygen produces nitric oxide (NO) and water as the only products. Write a balanced chemical equation for this reaction using the smallest possible whole number coefficients for the reactants and products.

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Video Transcript

Reaction of ammonia, NH3, with oxygen produces nitric oxide, NO, and water as the only products. Write a balanced chemical equation for this reaction using the smallest possible whole number coefficients for the reactants and products.

Let’s begin by identifying the reactants and products. The problem begins “reaction of ammonia with oxygen.” The keyword reaction indicates that ammonia and oxygen are the reactants. We should write the chemical formula of ammonia and oxygen on the left-hand side of the reaction arrow. Notice that oxygen is written as O subscript two. This is because oxygen exists as a diatomic molecule in its pure form. The reaction produces nitric oxide and water as the only products. We can therefore write NO and H2O on the products side of the reaction. Now we can begin balancing this chemical equation.

A chemical equation is balanced when the number of atoms of each element is the same on both sides of the reaction. To balance this chemical equation, we will start by identifying the elements involved in the reaction. In the equation, we can see the elements nitrogen, hydrogen, and oxygen. Now, we will count the number of atoms of each element on both sides of the reaction. We can draw a dashed line at the arrow to separate the reactants from the products.

Let’s begin by counting the nitrogen atoms. We see one atom of nitrogen on the reactants side and one atom of nitrogen on the products side. There are three atoms of hydrogen on the reactants side and two atoms of hydrogen on the products side. Lastly, let’s count the oxygen atoms. There are two oxygen atoms on the reactants side. And on the products side, we see one oxygen atom in each nitric oxide molecule and one oxygen atom in each water molecule. This gives us a total of two oxygen atoms on the products side.

Looking at our list, we can see that there’s one nitrogen atom on both sides of the reaction. This means that the nitrogen atoms are balanced. There are three hydrogen atoms on the reactants side and two hydrogen atoms on the products side. This means that the hydrogen atoms are unbalanced. And there are two oxygen atoms on both sides of the reaction, meaning that the oxygen atoms are balanced.

Now, we can add coefficients to the chemical equation in order to balance the reaction. Coefficients are numerical values that can be placed in front of any species on either side of the reaction. We only have one element that is out of balance, hydrogen. There are two options for how we could balance the hydrogen atoms. The first option is to place the fraction three-halves in front of the water molecule on the products side. If one water molecule has two hydrogen atoms, then three-halves water molecules will have three total hydrogen atoms. This would balance the hydrogen atoms.

However, placing a coefficient in front of a molecule affects all of the atoms in the molecule. This means that there are now three-halves oxygen atoms coming from the water molecules on the products side and that the total number of oxygen atoms on the products side is now five-halves. While we may have balanced the hydrogen atoms, we have unbalanced the oxygen atoms. And we are left trying to balance two oxygen atoms on the reactant side with five-halves oxygen atoms on the product side. This seems quite tricky.

The second way that we could try to balance the hydrogen atoms is to find the least common multiple of three and two. The least common multiple of three and two is six. This tells us that we will want to find a way to get six hydrogen atoms on the reactant side and six hydrogen atoms on the product side. On the reactant side, we could place a coefficient of two in front of the ammonia. If one ammonia molecule has three hydrogen atoms, then two ammonia molecules will have six hydrogen atoms.

We must also remember that placing a coefficient in front of a molecule affects all of the atoms in the molecule. If one ammonia molecule has one nitrogen atom, then two ammonia molecules will have two nitrogen atoms. On the products side, we could place a coefficient of three in front of the water molecule. If one water molecule has two hydrogen atoms, then three water molecules will have six total hydrogen atoms.

We will also remember that placing a coefficient of three in front of the water molecule also affects the number of oxygen atoms on the products side. There are now three oxygen atoms coming from the three water molecules and a total of four oxygen atoms on the products side. By placing coefficients to get six hydrogen atoms on both sides of the reaction, we have balanced the hydrogen atoms, but we have also unbalanced the nitrogen and oxygen atoms.

By using the method of the least common multiple, instead of the fraction three-halves, we have unbalanced more of the elements. However, we have kept all of the atom totals to whole numbers, making them easier to balance overall.

Now, we can try to rebalance the nitrogen atoms. There are two nitrogen atoms on the reactant side and one nitrogen atom on the product side. We can place the coefficient of two in front of the nitric oxide to give us two nitrogen atoms on the product side. We will also remember that placing a coefficient in front of the nitric oxide molecule will also affect the number of oxygen atoms on the product side. This means that there are now a total of five oxygen atoms on the product side. With two nitrogen atoms on both sides of the reaction, the nitrogen atoms are rebalanced. However, the oxygen atoms remain unbalanced.

We now need to decide what coefficient to place in front of the oxygen molecule in order for there to be five oxygen atoms on both sides of the reaction. Placing a coefficient of two-halves in front of the oxygen molecule will give us a total of five oxygen atoms on both the reactant and product sides, thus balancing the oxygen atoms.

The overall reaction is balanced. However, the question asked us to use whole number coefficients. We can get rid of the decimal two-halves and maintain a balanced chemical equation by multiplying all of the coefficients in the equation by two. This gives us a final balanced chemical equation of 4NH3 plus 5O2 react to produce 4NO plus 6H2O.

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