All chemical reactions (and therefore all chemical equations) must be balanced. Matter cannot be created or destroyed, so the products resulting from a reaction must match the participating reactants, even if they are arranged differently. Stoichiometry is the technique that chemists use to ensure that a chemical equation is perfectly balanced. Stoichiometry is half mathematical, half chemical, and focuses on the simple principle just outlined: the principle according to which matter is never destroyed or created during a reaction. See step 1 below to get started!
Steps
Part 1 of 3: Learning the Basics
Step 1. Learn to recognize the parts of a chemical equation
Stoichiometric calculations require an understanding of some basic principles of chemistry. The most important thing is the concept of the chemical equation. A chemical equation is basically a way to represent a chemical reaction in terms of letters, numbers and symbols. In all chemical reactions, one or more reactants react, combine, or otherwise transform to form one or more products. Think of reagents as the "base materials" and products as the "end result" of a chemical reaction. To represent a reaction with a chemical equation, starting from the left, we first write our reagents (separating them with the sign of addition), then we write the sign of equivalence (in simple problems, we usually use an arrow pointing to the right), finally we write the products (in the same way we wrote the reagents).
- For example, here is a chemical equation: HNO3 + KOH → KNO3 + H2O. This chemical equation tells us that two reactants, HNO3 and KOH combine to form two products, KNO3 and H2OR.
- Note that the arrow in the center of the equation is just one of the equivalence symbols used by chemists. Another often used symbol consists of two arrows arranged horizontally one above the other pointing in opposite directions. For the purposes of simple stoichiometry, it usually does not matter which equivalence symbol is used.
Step 2. Use the coefficients to specify the quantities of different molecules present in the equation
In the equation of the previous example, all the reactants and products were used in a ratio of 1: 1. This means we used one unit of each reagent to form one unit of each product. However, this is not always the case. Sometimes, for example, an equation contains more than one reactant or product, in fact it is not uncommon for each compound in the equation to be used more than once. This is represented using coefficients, i.e. integers next to the reactants or products. The coefficients specify the number of each molecule produced (or used) in the reaction.
For example, let's examine the equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. Note the "2" coefficient next to O2 and H2O. This equation tells us that a molecule of CH4 and two O2 form a CO2 and two H.2OR.
Step 3. You can "distribute" the products in the equation
Surely you are familiar with the distributive property of multiplication; a (b + c) = ab + ac. The same property is also substantially valid in chemical equations. If you multiply a sum by a numerical constant inside the equation, you get an equation which, although no longer expressed in simple terms, is still valid. In this case, you have to multiply each coefficient itself constant (but never the numbers written down, which express the amount of atoms within the single molecule). This technique can be useful in some advanced stoichiometric equations.
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For example, if we consider the equation of our example (CH4 + 2O2 → CO2 + 2H2O) and multiply by 2, we get 2CH4 + 4O2 → 2CO2 + 4H2O. In other words, multiply the coefficient of each molecule by 2, so that the molecules present in the equation are twice the initial equation. Since the original proportions are unchanged, this equation still holds.
It may be useful to think of molecules without coefficients as having an implicit coefficient of "1". Thus, in the original equation of our example, CH4 becomes 1CH4 and so on.
Part 2 of 3: Balancing an Equation with Stoichiometry
Do Stoichiometry Step 4 Step 1. Put the equation in writing
The techniques used to solve stoichiometry problems are similar to those used to solve math problems. In the case of all but the simplest chemical equations, this usually means that it is difficult, if not nearly impossible, to perform stoichiometric calculations in mind. So, to get started, write the equation (leaving enough space to do the calculations).
As an example, let's consider the equation: H.2SO4 + Fe → Fe2(SO4)3 + H2
Do Stoichiometry Step 5 Step 2. Check if the equation is balanced
Before starting the process of balancing an equation with stoichiometric calculations, which can take a long time, it is a good idea to quickly check whether the equation actually needs to be balanced. Since a chemical reaction can never create or destroy matter, a given equation is unbalanced if the number (and type) of atoms on each side of the equation does not match perfectly.
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Let's check if the equation of the example is balanced. To do this, we add the number of atoms of each type that we find on each side of the equation.
- To the left of the arrow, we have: 2 H, 1 S, 4 O, and 1 Fe.
- To the right of the arrow, we have: 2 Fe, 3 S, 12 O, and 2 H.
- The quantities of the atoms of iron, sulfur and oxygen are different, so the equation definitely is unbalanced. Stoichiometry will help us balance it!
Do Stoichiometry Step 6 Step 3. First, balance any complex (polyatomic) ions
If some polyatomic ion (consisting of more than one atom) appears in both sides of the equation in the reaction to be balanced, it is usually a good idea to start by balancing these in the same step. To balance the equation, multiply the coefficients of the corresponding molecules in one (or both) of the sides of the equation by whole numbers so that the ion, atom or functional group you need to balance is present in the same amount on both sides of the equation. 'equation.
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It is much easier to understand with an example. In our equation, H.2SO4 + Fe → Fe2(SO4)3 + H2, SO4 it is the only polyatomic ion present. Since it appears on both sides of the equation, we can balance the entire ion, rather than the individual atoms.
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There are 3 SOs4 to the right of the arrow and only 1 SW4 to the left. So to balance SO4, we would like to multiply the molecule on the left in the equation of which SO4 is part for 3, like this:
Step 3. H.2SO4 + Fe → Fe2(SO4)3 + H2
Do Stoichiometry Step 7 Step 4. Balance any metals
If the equation contains metallic elements, the next step will be to balance these. Multiply any metal atoms or metal-containing molecules by integer coefficients so that the metals appear on both sides of the equation in the same number. If you are not sure if atoms are metals, consult a periodic table: in general, metals are the elements to the left of the group (column) 12 / IIB except H, and the elements in the lower left of the "square" part to the right of the table.
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In our equation, 3H2SO4 + Fe → Fe2(SO4)3 + H2, Fe is the only metal, so this is what we will need to balance at this stage.
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We find 2 Fe on the right side of the equation and only 1 Fe on the left side, so we give the Fe on the left side of the equation the coefficient 2 to balance it. At this point, our equation becomes: 3H2SO4 +
Step 2. Fe → Fe2(SO4)3 + H2
Do Stoichiometry Step 8 Step 5. Balance the non-metallic elements (except oxygen and hydrogen)
In the next step, balance any non-metallic elements in the equation, with the exception of hydrogen and oxygen, which are generally balanced last. This part of the balancing process is a bit hazy, because the exact non-metallic elements in the equation vary greatly based on the type of reaction to be performed. For example, organic reactions can have large numbers of C, N, S, and P molecules that need to be balanced. Balance these atoms in the manner described above.
The equation of our example (3H2SO4 + 2Fe → Fe2(SO4)3 + H2) contains quantities of S, but we have already balanced it when we balanced the polyatomic ions of which they are a part. So we can skip this step. It is worth noting that many chemical equations do not require every single step of the balancing process described in this article to be performed.
Do Stoichiometry Step 9 Step 6. Balance the oxygen
In the next step, balance any oxygen atoms in the equation. In balancing the chemical equations, the O and H atoms are generally left at the end of the process. This is because they are likely to appear in more than one molecule present in both sides of the equation, which can make it difficult to know how to start before you have balanced the other parts of the equation.
Fortunately, in our equation, 3H2SO4 + 2Fe → Fe2(SO4)3 + H2, we have already balanced the oxygen previously, when we balanced the polyatomic ions.
Do Stoichiometry Step 10 Step 7. Balance the hydrogen
Finally, it ends the balancing process with any H atoms that may be left. Often, but obviously not always, this can mean associating a coefficient with a diatomic hydrogen molecule (H2) based on the number of Hs present on the other side of the equation.
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This is the case with the equation of our example, 3H2SO4 + 2Fe → Fe2(SO4)3 + H2.
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At this point, we have 6 H in the left side of the arrow and 2 H in the right side, so let's give the H.2 on the right side of the arrow the coefficient 3 to balance the number of H. At this point we find ourselves with 3H2SO4 + 2Fe → Fe2(SO4)3 +
Step 3. H.2
Do Stoichiometry Step 11 Step 8. Check if the equation is balanced
After you are done, you should go back and check if the equation is balanced. You can do this verification just as you did in the beginning, when you discovered that the equation was unbalanced: by adding all the atoms present in both sides of the equation and checking if they match.
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Let's check if our equation, 3H2SO4 + 2Fe → Fe2(SO4)3 + 3H2, is balanced.
- On the left we have: 6 H, 3 S, 12 O, and 2 Fe.
- To the right are: 2 Fe, 3 S, 12 O, and 6 H.
- You did! The equation is balanced.
Do Stoichiometry Step 12 Step 9. Always balance the equations by changing only the coefficients, and not the subscribed numbers
A common mistake, typical of students who are just starting to study chemistry, is to balance the equation by changing the inscribed numbers of the molecules in it, rather than the coefficients. In this way, the number of molecules involved in the reaction would not change, but the composition of the molecules themselves, generating a completely different reaction from the initial one. To be clear, when performing a stoichiometric calculation, you can only change the large numbers to the left of each molecule, but never the smaller ones written in between.
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Suppose we want to try to balance the Fe in our equation using this wrong approach. We could examine the equation studied just now (3H2SO4 + Fe → Fe2(SO4)3 + H2) and think: there are two Fe on the right and one on the left, so I'll have to replace the one on the left with Fe 2".
We can't do that, because that would change the reagent itself. The Fe2 it is not just Fe, but a completely different molecule. Furthermore, since iron is a metal, it can never be written in diatomic form (Fe2) because this would imply that it would be possible to find it in diatomic molecules, a condition in which some elements are found in the gaseous state (for example, H2, OR2, etc.), but not metals.
Part 3 of 3: Using Balanced Equations in Practical Applications
Do Stoichiometry Step 13 Step 1. Use stoichiometry for Part_1: _Locate_Reagent_Limiting_sub find the limiting reagent in a reaction
Balancing an equation is only the first step. For example, after balancing the equation with stoichiometry, it can be used to determine what the limiting reagent is. The limiting reactants are essentially the reactants that "run out" first: once they are used up, the reaction ends.
To find the limiting reactant of the equation just balanced, you have to multiply the quantity of each reactant (in moles) by the ratio between the product coefficient and the reactant coefficient. This allows you to find the amount of product that each reagent can produce: that reagent that produces the smallest amount of product is the limiting reagent
Do Stoichiometry Step 14 Step 2. Part_2: _Calculate_the_Theoretical_ Yield_sub Use stoichiometry to determine the amount of product generated
After you have balanced the equation and determined the limiting reactant, to try to understand what the product of your reaction will be, you just need to know how to use the answer obtained above to find your limiting reagent. This means that the quantity (in moles) of a given product is found by multiplying the quantity of the limiting reactant (in moles) by the ratio between the product coefficient and the reagent coefficient.
Do Stoichiometry Step 15 Step 3. Use the balanced equations to create the conversion factors of the reaction
A balanced equation contains the correct coefficients of each compound present in the reaction, information that can be used to convert virtually any quantity present in the reaction into another. It uses the coefficients of the compounds present in the reaction to set up a conversion system that allows you to calculate the arrival quantity (usually in moles or grams of product) from a starting quantity (usually in moles or grams of reagent).
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For example, let's use our above balanced equation (3H2SO4 + 2Fe → Fe2(SO4)3 + 3H2) to determine how many moles of Fe2(SO4)3 they are theoretically produced by a mole of 3H2SO4.
- Let's look at the coefficients of the balanced equation. There are 3 piers of H.2SO4 for each mole of Fe2(SO4)3. So, the conversion happens as follows:
- 1 mole of H2SO4 × (1 mole Fe2(SO4)3) / (3 moles H2SO4) = 0.33 moles of Fe2(SO4)3.
- Note that the quantities obtained are correct because the denominator of our conversion factor vanishes with the starting units of the product.
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