How do you write the balanced equation when 3.00 mol of C6H6 is burned in oxygen and it releases heat?

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To determine the stoichiometric ratios of molecules in a chemical equation (i.e., to balance it), it doesn’t matter how many mols you have of any specific molecule. “Burned in oxygen” means that you have a combustion reaction, and all combustion reactions use the same reactants and produce the same products: a carbon-containing compound, plus O2 yields H20 and CO2. For the combustion of C6H6, the unbalanced equation thus looks like this:

C6H6 + O2 -> CO2 + H20. Remember that molecular oxygen is always diatomic, so it will always be O2, not “O,” in its neutral state.

When you start balancing chemical equations, always begin with the elements that appear in the fewest number of molecules on each side to simplify your calculations. C only appears in one molecule on each side, so start with it. In the reactants, we have 6 C atoms, so put a coefficient of 6 in front of CO2:

C6H6 + O2 -> 6CO2 + H20.

Now balance for H. We have 6 H atoms in the reactants, so put a coefficient of 3 in front of H20, because this will multiply with the subscript “2” of H­20 to give 6 H’s in the products:

C6H6 + O2 -> 6CO2 + 3H20.

Now to balance the O atoms. You might notice that, after our first two steps, there is an odd number of O atoms in the products side of the equation, meaning that it is impossible to balance the equation by just adding a whole-number coefficient to the O2­ in the reactants. That’s ok; there is a solution around this. How many O atoms are on the right side? 15. So, ultimately, we want there also to be 15 O atoms on the left side. To get this, multiply O2 by the fraction 15/2:

C6H6 + (15/2)O2 -> 6CO2 + 3H20.

Now, in order to get rid of this fraction, multiply both sides of the chemical equation by 2. This trick works just the same way as it would in an algebra equation. Because the denominator of the fraction in front of O2 is “2,” multiplying by 2 will cancel it out:

(2)( C6H6 + (15/2)O2) -> (2)( 6CO2 + 3H20).

The result is:

2C6H6 + 15O2 -> 12CO2 + 6H20.

This is the balanced chemical equation.

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To write a balanced chemical equation, we need to know the chemical formula of all the species that are involved in the reaction. In other words, we need to list the reactants and products. When benzene (C6H6) is burned in oxygen, assuming complete combustion, carbon dioxide and water will be produced, along with heat. Thus, the reactants are benzene and oxygen, while the products are carbon dioxide, water, and heat.

Using all the chemical species, we can write the reaction as: C6H6 + O2 -> CO2 + H2O. Now, in a balanced chemical reaction, the atoms of any given species are the same on both sides of the equation. However, we can quickly check that this is not the case with the current equation. There are 6 atoms each of carbon and hydrogen on the reactant side, while there are only 1 atom of carbon and 2 atoms of hydrogen on the product side. To balance carbon atoms, we can have a coefficient of 6 for carbon dioxide, and similarly for balancing the hydrogen atoms, we can have a coefficient of 3 for water molecule. That is, C6H6 + O2 -> 6CO2 + 3H2O.

Now there are 6 carbon and 6 hydrogen atoms on each side of the equation. However, oxygen is still not balanced. There are 15 atoms of oxygen on the product side, while there are only 2 on the reactant side. To balance the oxygen atoms, we can add a coefficient of 15/2 (or 7.5) to the oxygen gas. That is, C6H6 + 15/2 O2 -> 6CO2 + 3H2O.

Now, one can count the number of atoms of each species on either side of the equation and check that the equation is balanced. If 3 moles of benzene is burned in oxygen, we can multiply the balanced equation with 3 and also add heat to the product side to get a balanced chemical equation. That is: 3C6H6 + 45/2 O2 -> 18CO2 + 9H2O + Heat. This is the required balanced chemical reaction. Hope this helps.

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