In Class Example

[Gabby Sigal 3H]

In the combustion of butane (C4H10) in a gas barbecue, balance the chemical equation C4H10+O2-- CO2+H20.

Can someone clarify why we multiply the O2 on the reactants by 13/2? I understand that after balancing the equation, we get 13 oxygen on the products side and thus we need 13 on the reactants side, but how does dividing 13/2 get us 13?

[Chloe Borja 2D]

If I remember correctly, we would use the stoichiometric coefficient 13/2 because the O2 on the reactants side has that little subscript. My interpretation might be a little difficult to follow, but I think about it as if we would need 13/2 or 6.5 O2 molecules on the reactants side, in order to balance out the products side. With reference to the number of Oxygen atoms on the reactants side, this would be 13 (6.5 x 2 atoms per O2 molecule).

Now, keeping 13/2 as a stoichiometric coefficient would be valid, but we would prefer to have whole number coefficients. Therefore, we would multiple both the reactants and products side by 2 in order to cancel out that denominator in 13/2.

I hope that helped a little.

[Chem_Mod]

I believe there may be some miscommunication here. After the initial balancing of the equation to account for the carbon and hydrogen atoms there are 13 oxygen atoms on the product side. This would require 13 oxygen on the reactants side as well. However oxygen is diatomic or for every oxygen molecule there are two oxygen atoms so you would need 13/2 oxygen molecules. However when balancing equations you do not use fractions of molecules so you then multiply the entire equation by 2 to get 13 oxygen molecules. 2C4H10 + 13O2 ---> 8CO2 + 10H2O

[005801694]

The reason you need 13/2 as the stoichiometric coefficient for oxygen on the reactants side is because you multiply the stoichiometric coefficient by the subscript of oxygen (number of atoms present) to get the total moles of oxygen. The chemical equation shows oxygen as O_2 so if you were to place the stoichiometric coefficient of 13 in front of O_2, you would actually get 26 moles of O since 13*2=26. You put 13/2 as the coefficient to account for the 2 atoms of oxygen in O_2 since (13/2)O_2 really means (13/2)*2= 13 total moles of oxygen. Since the stoichiometric coefficient should be a whole number, though, you now multiply both sides of the reaction by 2 to get 2C4H10+13O2->8CO2+10H20

[Matthew Vu 3C]

This is because on the reactants side, it is O2, no O. By multiplying by 13/2, the 2's cancel out, so you're left with 13 oxygen atoms. However, we want integers as our stoichiometric coefficients, which is why we multiply all the reactants and products by 2 at the end.

[Justin_Choo_3J]

The value 13/2 was used to balance the reactant portion of the equation with the products. However, a general rule of thumb is that the coefficients of both sides of the equation must be whole numbers for the sake of providing a clear and concise ratio between the reactants and products in question. Multiplying the aforementioned value by 2 would provide the whole number, 13.

[Christina Gigoux 1D]

The coefficient of 13/2 was simply an intermediate step based on the order in which we balanced the equation. After adding stoichiometric coefficients so that the C and H were balanced, the products contained 13 O, so we used the coefficient of 13/2 for O2 because 13/2 x 2 = 13. However, stoichiometric coefficients must always be the lowest possible whole number, and 13/2 is not a whole number. We then multiply all of the coefficients on both the reactants and products sides by the lowest possible value in order to make all of the coefficients whole numbers, and in this case we multiplied by a factor of 2.

[Martha Avila 1I]

We multiply the O2 on the reactants side by essentially 6.5 so that it would equal the same amount of O2 on the products side which is 13. However because it is not ideal to multiply by 6.5 we then are going to multiply by 13/2 in order to avoid dealing with a decimal. Once we reach this point we want to have a whole number as the stoichiometric coefficient so we multiply the whole equation by the denominator which is 2. This will turn our coefficient into 13 instead of 13/2. Since we multiplied this molecule by 2 we also have to do the same to the entire equation which gives you a final balanced equation of 2C4H10 +13 O2 → 8CO2 + 10H2O. Hope this helps.

[Sarah Hong 2K]

By multiplying O2 by the coefficient 13/2, the 2's cancel out leaving 13 moles of oxygen. However you can not have a fraction as a stoichiometric coefficient so you must multiply EVERYTHING by 2 to cancel out the denominator of 13/2. So once you multiply the whole equation by 2, you will get 2C4H10 + 13O2= 8CO2 + 10H2O.

[Lindsey Walter 3E]

We would need 13/2 O2 because that is what would make the equation balanced. There are 13 oxygen atoms on the product side, therefore we need 13 on the reactant side. Since the only oxygen atoms on the reactant side are found in O2, we must have 13/2 because, for every O2 that we have, we get two oxygen atoms. Therefore 13 O2 would actually be 26 oxygen atoms, which is too many for a balanced equation. However, even though 13/2 is the correct answer, it is not stoichiometrically correct and we need a whole number. To get a whole number we would multiply EVERY coefficient in the reaction by 2, so that 13/2 became 13. However, all of the other coefficients are doubled too, so the reaction is still balanced.