Balancing Chemical Equations - Online Tutor, Practice Problems & Exam Prep

When approaching the balancing of chemical equations, it is crucial to ensure that both the type and number of atoms are equal on each side of the reaction arrow. Coefficients, which are the numbers placed before the compounds in the equation, are used to achieve this balance. These coefficients must be distributed across the atoms they precede to determine the total number of atoms present. For instance, a coefficient of 2 before a molecule containing hydrogen would mean that there are effectively twice as many hydrogen atoms. Achieving a balanced chemical equation means that for each element, the quantity of atoms in the reactants equals the quantity in the products, maintaining the law of conservation of mass. This fundamental concept is essential for students to grasp as it forms the basis of stoichiometry and is a key skill in the field of chemistry.

Balancing Chemical Equations

In this series of videos, we're going to take a look at balancing chemical equations. When balancing, always make sure the type and number of atoms on both sides of the arrow are equal. We're going to say in a balanced equation, the numbers that are in red are referred to as the coefficients. So we have two one and two as the coefficients for this balanced equation.

And when we're talking about type and number of atoms, we'd say that these coefficients get distributed. So this two would get distributed to this hydrogen. There's already two of them, so 2×2=4 hydrogens. This is just a one. So one distributed is 1×2=2 oxygens. On the other side, the two gets distributed, so it'd be two times the two hydrogens Would give me 4 hydrogens. But there's also one oxygen here, so 2×1=2 oxygens.

So the types of atoms are the same on both sides. I have hydrogen and hydrogen, oxygen and oxygen, and then the numbers of each are equal on both sides, 4 hydrogens, 4 hydrogens, 2 oxygens, 2 oxygens. This is how we ensure a chemical equation is balanced. The types and numbers of each atoms are the same on both sides of the arrow.

Now click on the next video and let's take a look at an example question where we go more refined at our approach to balancing a chemical equation.

Balancing Chemical Equations Example 1

Here it says to write the balanced equation for the following by inserting the correct coefficients in the blanks. So in this reaction we have C₄H₁₀O reacting with O₂ to give us water and CO₂ as products. Now here's step one says that we need to set up a list for the elements that are reactants, so those before the arrow, and another list for the elements that are products, those that come after the arrow.

So if we take a look here, we're going to say that we have C₄H₁₀O₂. That gives us 4 carbons, 10 hydrogens and two oxygens. Now on the right side of the arrow we have H₂O and CO₂. Here we'd say that we have one carbon, 2 hydrogens and then look at the number of oxygens, one here, but then two more here. That's three. So we've created our list.

Step two, start from the top and going down. Both lists determine how many of each element is present. We've done that. And then we're going to say step three. Start, start from the top and going down. Both lists begin balancing each element to ensure they match. Alright, you won't worry about that last line until we need to do it all right?

So if we look, we have 4 carbons on the left side, but only one on the right side. So that means we'd have to put a coefficient of four here. That four get distributed to the carbon, so 4 * 1 carbon gives us four, but it would also be distributed to the oxygen. So 4 * 2 is 8 + 1 more oxygen over here. That's nine carbons are balanced. Let's keep going down the list.

Hydrogens. We have 10 versus 2. I'd have to put a 5 here. 5 * 2 gives me 10 hydrogens, but then five times the one Oxygen is five oxygens, but then we have 8 over here so 5 + 8 gives me 13. My hydrogens are balanced but my oxygens are not. So now we have to think about what number can I put here with O₂ to give me 13. If I put a six there, that'll only give me 12. Not enough. If I put A7 there, that would give me 14 too much. So I need a number between 6:00 and 7:00 that would give me 13.

The number is 6.5. A 6.5 * 2 would give me 13, and here they'd just be a one. This represents my balance equation. But we have an issue. We have a basically a decimal .5 there. We're going to say in those instances where we have .5, we're going to say sometimes you may have a decimal or a fraction as a coefficient, so you must multiply the equation by two.

So I'm going to come back up here. I'm going to multiply this entire equation times 2. So all my coefficients get doubled. So now that's going to become two, this is going to become 13, This is going to become ten. And then here my CO₂ becomes 8. So then we're going to come back, We're going to plus and then plus, it's the arrow and then we're going to lug in what we know.

So we have 2C₄H₁₀ + 13O₂ will give me 10H₂O and 8CO₂. This would represent my balanced equation. And here we just put this as a gas, This is a gas, this is a liquid and this is a gas. So this would represent my completely balanced chemical equation, right? So this is the approach that you have to take whenever it comes to balancing any type of chemical equation presented before you on an exam, homework, wherever you may see it.

Write the balanced equation for the following by inserting the correct coefficients in the blanks.

___ Al₄C₃ (s) + ___ H₂O (l) → ___ Al₂O₃ (s) + ___ CH₄ (g)

Here in this practice question, it says, write the balanced equation for the following by inserting the correct coefficients in the blanks. Alright. So remember, what we do first is we make a list of the reactants on one side and a list of products on the other. On the left side, we have aluminum, carbon, hydrogen, and oxygen. It doesn't matter the order of the elements on the product side; we need our lists to match, so it's also in the order of aluminum, carbon, hydrogen, and oxygen. Now on the reactant side we have 4 aluminums, we have 3 carbons, 2 hydrogens, and 1 oxygen. On the product side, we have 2 aluminums, 1 carbon, 4 hydrogens, and 3 oxygens. We can see that our lists don't match, so we have to put coefficients in front of each of these compounds. So if we look on the reactant side, we have 4 aluminums, but only 2 here. So I'd have to put a coefficient of 2 here. That 2 gets distributed to the aluminum to give us 4, but it also gets distributed to the oxygen to give us 6. Now move on to carbons. Here we have 3 carbons, here we only have 1. I'm going to place a 3 here to give us 3 carbons, but then it also gets distributed to the hydrogens to give me 12 hydrogens. Then for hydrogens, this is 12 but this is only 2, so I put a 6 here. That's going to give me 12 hydrogens and 1 oxygen product. My lists match, so my reaction is balanced. And if you want, you can put a coefficient of 1 right there. Just realize that if you don't see a coefficient there, you always assume that it's one that's invisible. Okay? So here our coefficients would be 1, 6, 2, and 3 for a balanced chemical reaction.

Determine the total sum of the coefficients after balancing the following equation.

___ C₂H₅SH (g) + ___ O₂ (g) → ___ CO₂ (g) + ___ H₂O (l) + ___ SO₂ (g)

In this practice question, it says, determine the total sum of the coefficients after balancing the following equation. Alright. So this one's a little bit different from what we're accustomed to seeing. Here, they're basically asking us to find all the coefficients for each compound and then add them up together because we want the sum, the total sum of all the coefficients. Alright. So we're going to start out with our list. We're going to say we have carbon, hydrogen, sulfur, and oxygen. On the other side, the list matches carbon, hydrogen, sulfur, and oxygen. Now we're going to say we have 2 carbons, 5+1, 6 hydrogens, 1 sulfur, and 2 oxygens. On the other side, we have 1 carbon, 2 hydrogens, 1 sulfur, and then we have what? 2 oxygens, 3 oxygens, 5 oxygens, 5 oxygens here. Alright. So let's balance this out. The number of carbons don't match. Here we have a 2 and here we have a 1. So we're going to place a coefficient of 2 right here. That 2 gets distributed to not only the carbon but also the oxygen. So we're going to have 2 carbons and then 2 times 2 is 4 oxygens, 5 oxygens, and then 7 oxygens. Next, balance out your hydrogens. Here this is 6, but here this is only 2, so I'm going to put a 3 here. So 3 times 2 gives me 6 hydrogens, but then we have 3 oxygens here plus another 4 here, so that's 7, plus another 2 is 9 oxygens. Don’t worry, we're just going down the list, we’re going to eventually get to the oxygens. Sulfurs are the same, both are 1 and 1. So let's not do anything to that. Then here we have 2 oxygens and here we have 9 oxygens. So we have to think of a number to multiply 2 by to get to 9. We can’t do 4 because 4 times 2 gives us 8. We can’t do 5 because 5 times 2 gives us 10. The number that we have to place here is 4.5. 4.5 times 2 would give me 9. But remember, you're not allowed to have fractions or decimals as coefficients. They have to be whole numbers. In these situations, remember, all we do is we multiply this by 2. So all the coefficients will get multiplied by 2. So when we do that, we’re going to have 2, 9, 4, 6, and 2 as the coefficients for each one of these compounds. K. So then 2 O2 gas gives me 4 CO2 gas + 6 H2O liquid + 2 SO2 gas. So our coefficients would be 2, 9, 4, 6, and 2. This is not our final answer because, remember, we have to add them all up. We need the total sum of all the coefficients. So 2 + 9 gives me 11, 11 + 4 equals 15, plus 6 gives me 21, plus this final 2 gives me 23 as the total sum for all the coefficients in this balanced equation. So 23 would be our final answer.

Determine the balanced chemical equation when ethanol, C₂H₆O is ignited in the presence of air.

• A. C₂H₆O + 4 H₂O → 5 CO₂ + 7 CO 
• B. C₂H₆O + 2 O₃ → 2 CO₂ + 3 H₂O
• C. C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
• D. C₂H₆O + 3 O₂ → 5 CO₂ + 4 H₂O

Here’s what students ask on this topic:

How do you balance chemical equations?

Balancing chemical equations is a process of ensuring that the number of atoms for each element is the same on both the reactant and product sides of the equation. Here's a step-by-step guide:

1. Write down the unbalanced equation.
2. List the number of atoms of each element present in the reactants and products.
3. Start by balancing elements that appear in only one reactant and one product.
4. Balance elements that are in multiple compounds, saving hydrogen and oxygen for last.
5. Adjust coefficients (the numbers in front of compounds) to get the same number of atoms of each element on both sides. Remember, you can only change coefficients, not subscripts in the chemical formulas.
6. As you change coefficients, continually update the count of atoms for each element.
7. Check your work by ensuring that all elements are balanced, meaning the atom count is equal on both sides.
8. Finally, make sure that all coefficients are in the lowest possible ratio.

For example, to balance the equation:

$H_2+O_2\to H_{2}O$

you would start by putting a coefficient of 2 in front of H₂O to balance the hydrogen atoms, resulting in:

$H_2+O_2\to 2H_{2}O$

Then, to balance the oxygen, you would put a coefficient of 2 in front of the O₂, giving:

$2H_2+O_2\to 2H_{2}O$

Why do we balance chemical equations?

Balancing chemical equations is essential because it reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means that the number of atoms of each element must be the same on both the reactant side and the product side of the equation. By balancing the equation, we ensure that the equation accurately represents the stoichiometry of the reaction, which is the quantitative relationship between reactants and products.

Balancing chemical equations allows chemists to predict the amounts of substances consumed and produced in a reaction, which is crucial for experiments and industrial processes. It helps in calculating the correct proportions of reactants needed and the expected yield of products, which is critical for efficiency and cost-effectiveness in chemical manufacturing. Additionally, it is necessary for understanding reaction mechanisms and for the safe handling of chemicals, as unbalanced reactions may lead to incorrect conclusions about the nature of the reaction taking place.

Why do we adjust coefficients when balancing chemical equations and not subscripts?

When balancing chemical equations, we adjust coefficients instead of subscripts because subscripts are part of the chemical formula that indicate the number of atoms of each element in a molecule, which defines the substance's identity. Changing a subscript would change the actual compound. For example, $H_{2}O$ represents water, but altering the subscript to make it $H_2{O}_2$ would turn it into hydrogen peroxide, a completely different substance.

Coefficients, on the other hand, are numbers placed in front of the chemical formulas to indicate the number of molecules or moles of that substance. Adjusting coefficients ensures that the same number of each type of atom appears on both sides of the equation, which is necessary to satisfy the Law of Conservation of Mass. This law states that in a chemical reaction, matter is neither created nor destroyed. By changing coefficients, we can balance the equation without altering the identity of the substances involved.