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## MCAT

### Course: MCAT > Unit 8

Lesson 26: Balancing chemical equations- Balancing chemical equations 1
- Balancing chemical equations 2
- Balancing chemical equations
- Balancing more complex chemical equations
- Visually understanding balancing chemical equations
- Balancing another combustion reaction
- Balancing chemical equation with substitution
- Balancing Chemical Equations Intuition

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# Balancing another combustion reaction

Balancing the combustion reaction of ethane, C₂H₆ .

## Want to join the conversation?

- Do the letters in parentheses make a difference? i.e.: (g)(77 votes)
- Yes, they tell you what state the substance is in. That matters a great deal in how the substance behaves in general and how it participates in the reaction in particular.(97 votes)

- How are the oxygens balanced at the end? It looks like you end up with 7 on one side and 14 on the other.(22 votes)
- As another note, bear in mind that in the final "balanced" reaction, the presenter messed up and accidentally wrote C2H2, not C2H6.(22 votes)

- i did the reaction on paper before he started and got 1 C2H2 + 8 O2 = 2 Co2 + 6 H2O. what did i do wrong? it's all balanced and everything.

{answered}(11 votes)- The equation is actually....

C2H6 + O2 ===> Co2 + H2O.

Whereas you have written C2H2 in your equation. Hence, a different result.(10 votes)

- So when someone says "CO2" it is C*O2 and not (CO) squared, correct? (I know it isn't a mathematical equation, just relating)(4 votes)
- Yes you are correct @Ryan McHale, the numbers that are in between the molecular formula, or in other words, in between the "letters" are the number of that particular atom that is behind the number, and it's not actually written just like that... It is in the subscript, like the opposite area of where you write the Indices in Mathematics which means a little bit below the letter.

The letter (element) that has a number written in the subscript of it will be the number of atoms of that particular element, and the number that is written normally**before**the**whole**molecule, is the number of that particular molecule in the equation! So for`CO2`

, there are 1 Carbon, and 2 Oxygen atoms and for`2 CO2`

, there are 2 molecules of Carbon dioxide, so altogether we would have 2 Carbon, and 4 Oxygen atoms. Hope this helps!(5 votes)

- did sal just change the subscript value of the hydrogen in the rewritten reaction @4:41(4 votes)
- Sal wrote C2H2 by mistake. he actually maent C2H6(3 votes)

- You can multiply an element by a decimal to balance an equation?(4 votes)
- Only if the final stoceometric ratio is in whole numbers.(4 votes)

- When I tried to balance this reaction on my own, I got:

4C2H6 + 14O2 --> 8CO2 + 12H20

If I wrote that instead of what Sal had at4:48, would I still get credit? Thanks!(2 votes)- You may get partial credit but it isn’t correct because you can divide all your coefficients by 2 to get what Sal got. The correct balanced equation always has the lowest whole number coefficients.(5 votes)

- At4:20their are fourteen oxygen but then at the last oxygen he only has twelve. Did he do something wrong OR do I just not understand.(2 votes)
- If you mean for the very last equation, there are 14 oxygens atoms on both sides. The reactant side has 7O2 which is 14 oxygen atoms (7x2=14). The product side has 4CO2 which is 8 oxygen atoms (4x2=8) and 6H2O which is 6 oxygen atoms for a total of 14 oxygen atoms (8+6 =14).

Hope that helps.(5 votes)

- Do we need to incorporate the letters in parentheses i.e. H2O (L)?(2 votes)
- Yes sir. We also use lower case l, which when written, has a small curve to it. These are written smaller than the rest of the formula.(5 votes)

- what is a combustion reaction(2 votes)
- A combustion reaction is when a
*fuel*, usually a hydrocarbon, reacts with excess oxygen to form carbon dioxide gas and water vapor.

Example: CH4 + 202 --> CO2 +2H2O(3 votes)

## Video transcript

- All right now we have
another combustion reaction. Instead of ethylene, we
now have ethane, C2H6, has two carbons and six hydrogen atoms in each molecule of ethane, and it is reacting. It's ethane gas, it is
reacting with molecular oxygen in gaseous form and they combust to form carbon dioxide gas and liquid water, and like we've seen in previous examples, this chemical equation is not balanced. How can we tell? Well here on the left hand
side we have two carbons. Here on the right hand
side we have one carbon. Here on the left hand side
we have six hydrogens. Here on the right hand side
we only have two hydrogens. Here on the left hand
side we have two oxygens. On the right hand side we
have two plus, three oxygens, so none of the elements here are balanced. But like we did in the
example of ethylene, whenever you see this
where you have several somewhat complex molecules involved, it's good to save the element that is in a molecule by itself for last, because you can just tweak this to change the number of oxygens without it having any other side effects on the number of carbons or hydrogens. So what I'm going to do is
I'm going to first balance, like we've done before, the carbons and the hydrogens, which are going to have
implications on the oxygens because if I change the number here, it's going to change
the number of oxygens. If I change the number here it's going to change the number of oxygens. But lucky for me I have
this dioxygen molecule on the left hand side that
I can just tweak at the end to balance the entire chemical equation. So let's start with, you know last time we started with carbon, let's
start with hydrogen this time. Just for kicks. So over here I have six hydrogens on the left hand side, of
the entire left hand side I only have six hydrogen atoms. On the right hand side I
only have two right now, so if I want to have six I would
multiply these two by three so now I have three water molecules, each of them have two hydrogen atoms, so I'm going to have
six, six hydrogen atoms on the right hand side. Fair enough. Now let's go to the carbon. Remember I'm saving oxygen for last. Carbon on the left hand
side I have two carbons. How many carbons do I have
on the right hand side? Well right now I only have one, but I can change that very easily. Instead of having one
molecule of carbon dioxide, I can have two molecules
of carbon dioxide. And so now my carbons are balanced, two carbons, two carbons. And now let's go to the oxygens. So right now, right now on,
I'll do this in a mauve color, right now on the left hand
side I have two oxygens, but on the right hand side what do I have? Let's see, I have two times two, so this right over here is four oxygens, and then I have three water molecules, each of them have one oxygen atom, so three times one, so
this is going to be three right over here. So on the entire right hand
side I have seven oxygen atoms and on the left I only have two. So what can I do here? What can multiply by two to get to seven? Two times what is equal to seven? Well two times three-and-a-half
is equal to seven. So two times three-and-a-half
is equal to seven. Remember I have two here,
I'm saying two times something is equal to seven. I want to get to the four plus the three. Well I multiply it by three point five, and now I have seven, seven oxygen atoms on both sides of my chemical equation. But like we've seen in previous videos, it is not standard to just
leave a three-and-a-half here, it's kind of this weird notion of three-and-a-half molecules, we like to have whole numbers here. So how do we make sure
we have all whole number coefficients in front of our molecules? Well we could just
multiply everything by two, then this thing is
going to become a seven, this thing is going to become a two, this is going to become a four, this is going to become a six, so let's just do that. I'll write the whole reaction over again. So I have my ethane, and I won't write actually what state it's
in just to save some time, plus some molecular oxygen. They combust, they yield, so these are the reactants, the products are carbon dioxide gas and liquid water, and liquid water. So let's see, you multiply,
if you say that there was a one out here before, we're going to multiply
that by two to get two. We had a three point five
here, multiply that by two, you get a seven here. We had a two right over
here, multiply that by two you get a four. Once again I'm just multiplying
all the coefficients by two just like you did
in algebraic equations in your algebra classes. And then finally three times two is six. And we're all balanced. We were balanced here but we didn't have whole number coefficients. Multiplying everything by two gave us the whole number coefficients
and we are much happier.