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Taking a dive into the world of chemical equations? These problems can seem tricky at a glance, but they’re easy to figure out once you learn the basic steps and rules to balancing them. Not to worry; we’ll walk you through exactly how to figure out just about any problem, no matter how many atoms and molecules you're working with. Dealing with especially complex equations? We’ve got you covered there, too—scroll to section 2 for a handy tutorial on solving trickier equations with an algebraic balance.

Method 1
Method 1 of 2:

Doing a Traditional Balance

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  1. Write down your given equation. For this example, you will use:
    • C 3 H 8 + O 2 --> H 2 O + CO 2
    • This reaction occurs when propane (C 3 H 8 ) is burned in the presence of oxygen to produce water and carbon dioxide.
  2. Do this for each side of the equation. Look at the subscripts next to each atom to find the number of atoms in the equation. When writing it out, it's a good idea to connect it back to the original equation, noting how each element appears. [1]
    • For example, you have 3 oxygen atoms on the right side, but that total results from addition.
    • Left side: 3 carbon (C3), 8 hydrogen (H8) and 2 oxygen (O2).
    • Right side: 1 carbon (C), 2 hydrogen (H2) and 3 oxygen (O + O2).
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  3. Hydrogen and oxygen are both common in molecules, so it's likely that you'll have them on both sides of your equation. It's best to balance them last. [2]
    • You'll need to recount your atoms before balancing the hydrogen and oxygen, as you'll likely need to use coefficients to balance the other atoms in the equation.
  4. If you have more than one element left to balance, select the element that appears in only a single molecule of reactants and in only a single molecule of products. This means that you will need to balance the carbon atoms first.
  5. Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation.
    • C 3 H 8 + O 2 --> H 2 O + 3 CO 2
    • The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3 on the left side indicates 3 carbon atoms.
    • In a chemical equation, you can change coefficients, but you must never alter the subscripts.
  6. Since you have balanced all atoms besides the hydrogen and oxygen, you can address the hydrogen atoms. You have 8 on the left side. So you'll need 8 on the right side. Use a coefficient to achieve this. [3]
    • C 3 H 8 + O 2 --> 4 H 2 O + 3CO 2
    • On the right side, you now added a 4 as the coefficient because the subscript showed that you already had 2 hydrogen atoms.
    • When you multiply the coefficient 4 times by the subscript 2, you end up with 8.
  7. Remember to account for the coefficients that you've used to balance out the other atoms. Because you've added coefficients to the molecules on the right side of the equation, the number of oxygen atoms has changed. You now have 4 oxygen atoms in the water molecules and 6 oxygen atoms in the carbon dioxide molecules. That makes a total of 10 oxygen atoms. [4]
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Method 2
Method 2 of 2:

Completing an Algebraic Balance

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This method, also known as Bottomley's method, is especially useful for more complex reactions, although it does take a bit longer.

  1. For this example, we will use:
    • PCl 5 + H 2 O --> H 3 PO 4 + HCl
    • a PCl 5 + b H 2 O --> c H 3 PO 4 + d HCl
  2. [5]
    • a PCl 5 + b H 2 O --> c H 3 PO 4 + d HCl
    • On the left side, there are 2 b atoms of hydrogen (2 for every molecule of H 2 O), while on the right side, there are 3 c + d atoms of hydrogen (3 for every molecule of H 3 PO 4 and 1 for every molecule of HCl). Since the number of atoms of hydrogen has to be equal on both sides, 2 b must be equal to 3 c + d .
    • Do this for each element.
      • P: a = c
      • Cl: 5 a = d
      • H: 2 b =3 c + d
  3. Since there are more variables than equations, there are multiple solutions. You must find the one where every variable is in its smallest, non-fractional form.
    • To quickly do this, take one variable and assign a value to it. Let's make a = 1. Then start solving the system of equations to get the following values:
    • Since P: a = c, we know that c = 1.
    • Since Cl: 5a = d, we know that d = 5
    • Since H: 2b = 3c + d, we can calculate b like this:
      • 2b = 3(1) + 5
      • 2b = 3 + 5
      • 2b = 8
      • b=4
    • This shows us the values are as follows:
      • a = 1
      • b = 4
      • c = 1
      • d = 5
    • If the value you assigned returns fractional values, just multiply all values by the least common multiple (LCM) of the denominators to get rid of the fractions. If there is only one fraction, multiply all values by that values denominator.
    • If the value you assigned returns coefficients that have a greatest common factor (GCF), simplify the chemical equation by dividing each value by the GCF.
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Community Q&A

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  • Question
    How did the oxygen became 7 when 3x2 is 6? Do you always add one?
    Community Answer
    You get 3x2=6 oxygen atoms in the carbon dioxide, but at this stage there is also 1 oxygen atom in the water. So, 6+1=7.
  • Question
    How can I balance NaCl + AgNO3 -> AgCl + NaNO3?
    Community Answer
    This equation is already balanced when using the first method (traditional balance) in the article.
  • Question
    What causes the color of copper sulphate to change when an iron nail is dipped into it?
    Community Answer
    When iron (Fe) and copper sulphate (CuSO4) solution react, they undergo a single displacement reaction, also known as a substitution reaction, to form solid copper (Cu) and aqueous iron sulphate (FeSO4). The iron can be solid or aqueous but the copper sulphate must be aqueous in order to facilitate the reaction.
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      Tips

      • Remember to simplify! If all of your coefficients can be divided by the same number, do so to get the simplest result.
      • If you're stuck, you can type the equation into the online balancer to balance it. Just remember that you won't have access to an online balancer when you're taking an exam, so don't become dependent on it.
      • To get rid of fractions, multiply the entire equation (both the left and right sides) by the number in the denominator of your fraction.
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      Warnings

      • During the balancing process, you may use fractions to assist you, but the equation is not balanced as long as there are still coefficients using fractions. You never make half of a molecule or half of an atom in a chemical reaction.
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      About This Article

      Article Summary X

      To balance a chemical equation, first write out your given formula with the reactants on the left of the arrow and the products on the right. For example, your equation should look something like "H2 + O2 → H2O." Count the number of atoms in each element on each side of the equation and list them under that side. For the equation H2 + O2 → H2O, there are 2 hydrogen atoms being added to 2 oxygen atoms on the left, so you would write "H=2" and "O=2" under the left side. There are 2 hydrogen atoms and 1 oxygen atom on the right, so you would write "H=2" and "O=1" under the right side. Since the number of atoms in each element isn't identical on both sides, the equation is not balanced. To balance the equation, you'll need to add coefficients to change the number of atoms on one side to match the other. For the equation H2 + O2 → H2O, you would add the coefficient 2 before H2O on the right side so that there are 2 oxygen atoms on each side of the equation, like H2 + O2 → 2H2O. However, subscripts can't be changed and are always multiplied by the coefficient, which means there are now 4 hydrogen atoms on the right side of the equation and only 2 hydrogen atoms on the left side. To balance this, add the coefficient 2 before H2 on the left side of the equation so there are 4 hydrogen atoms on each side, like 2H2 + O2 → 2H2O. Now the number of atoms in each element is the same on both sides of the equation, so the equation is balanced. Remember that if there's no coefficient in front of an element, it's assumed that the coefficient is 1. To learn how to balance chemical equations algebraically, scroll down!

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