What is the balanced equation is the reactant equal to the product?

A chemical equation is the chemical formula that provides the information of the elements and molecules that are reacting as well as the molecules that are being produced from that reaction. The Law of Conservation of Mass states that the mass of the reactants must balance the mass of the products. To balance a chemical equation, the atoms of both the elements and molecules on the reactant side (left side) and product side (right side) must be equal to each other.

In this instructable, you will understand and learn how to balance a chemical equation.

This instructable should take no longer than ten minutes.

The unbalanced chemical equation is given to you.

Aluminum reacts with oxygen to produce aluminum oxide.

Rewrite the equation as shown above.

First, identify the elements on the reactant side(left side) and the elements of the compound are on the product side (right side).

make a list of all of the elements on each side under the equation for both the reactants and products as shown above
Under the reactant’s side, list Al and O
under the product’s side, list Al and O

An atom is the smallest component of an element that contains chemical properties of that element. The atom of each element’s contains the protons, neutrons, and electrons of that element.

The list made of each element on both the reactant and product side will further help you identify the number of atoms each element contains.

Next to each element of the list, put the number of atoms that are in each of the elements.

on the reactants side, next to Al, put 1
next to O, put 2
apply these same rules to each element on the product side

Notice how the number of atoms next to each element is different from the number of atoms next to that same element on the product side.

In order to balance the chemical equation, you need to make sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. In order make both sides equal, you will need to multiply the number of atoms in each element until both sides are equal.

As shown above, the multiplication of the atoms on the reactant side will affect both elements on the product side.

After you have multiplied the number of atoms of each element until both sides are equal, you will put the number, the coefficient, of how much you multiplied the element by and place in front of that element or compound in the equation as shown above.

On the product side, although both elements did not get multiplied, still place the number that was multiplied as the coefficient in front of the compound.

After you have placed the coefficients in front of the molecules, make the list of elements again and check to see if multiplying the coefficient with the subscript will give you atoms equal on both the reactant and product sides.

If they are not equal, rework your multiplication.

After you have reworked your multiplication, make the list of elements again to check to make sure the equation is balanced. If both sides are equal, you have now balanced the chemical equation!

You surely have heard that matter can not be created or destroyed. This law applies to chemical reactions. In a chemical reaction, atoms can not be created or destroyed; they simply rearrange themselves to form new products. This law has an effect on the coefficients of a chemical equation. All of the atoms that were present at the beginning of the reaction as reactants also need to be present at the end of the reaction as products. A chemical equation that is written so this is true is said to be balanced.

Was the chemical equation previously discussed balanced? Let's take a closer look:

If this equation is balanced, the same atoms (in number and identity) will be present as reactants and products.

To see if the equation is balanced, we can follow two steps:

Step 1: Break each molecule up into the individual atoms. Count the number of each type of atom in each type of molecule.

Fill in the total number of atoms present for each type of molecule.

Step 2: Count the number of atoms of each type on each side of the equation (for the reactants and for the products). If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced.

Fill in the following table for this reaction:

Is this equation balanced?

Yes

Click on image below to see a pictorial representation of this. Clicking on this mouse will reset the image.

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Chemistry is the study of chemical reactions, processes in which reactants are converted to products. Chemical equations are a shorthand way of representing these reactions. They are always written:

Reactant(s)

Product(s)

The arrow in this shorthand notation can be thought of as meaning "forms" or "yields".

Ethylene (C2H4) is a colorless gas that causes fruit to ripen when exposed to it. This occurs because ethylene reacts with the oxygen gas in the air to form carbon dioxide and water. These products help speed the ripening process of fruit.

This chemical reaction can be expressed as:

There are several important features of a chemical equation; moving your mouse over the above chemical equation will point these out.

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Balance a chemical equation
Interpret the information conveyed by a balanced chemical equation

balanced chemical equation: a chemical equation that shows equal numbers of atoms of each kind in the products and the reactants.

stoichiometry: the study of the quantitative relations between amounts of reactants and products.

stoichiometric coefficients: the multiplying numbers assigned to the species in a chemical equation in order to balance the equation.

In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the equilibrium constants because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants.

There are several important rules for balancing an equation:

An equation can be balanced only by adjusting the coefficients.
The equation must include only the reactants and products that participate in the reaction.
Never change the equation in order to balance it.
If an element occurs in only one compound on each side of the equation, try balancing this element first.
When one element exists as a free element, balance this element last.

Example \(\PageIndex{1}\):

\[H_2\; (g) + O_2 \; (g) \rightleftharpoons H_2O \; (l) \nonumber \]

Because both reactants are in their elemental forms, they can be balanced in either order. Consider oxygen first. There are two atoms on the left and one on the right. Multiply the right by 2	\[H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l) \nonumber \]
Next, balance hydrogen. There are 4 atoms on the right, and only 2 atoms on the left. Multiply the hydrogen on left by 2	\[2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)\nonumber \]
Check the stoichiometry. Hydrogen: on the left, 2 x 2 = 4; on right 2 x 2= 4. Oxygen: on the left: 1 x 2 = 2; on the right 2 x 1 = 2 . All atoms balance, so the equation is properly balanced.	\[2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)\nonumber \]

Example \(\PageIndex{2}\):

\[Al \; (s) + MnSO_4 \; (aq) \rightleftharpoons Al_2(SO_4)_3 + Mn ; (s) \nonumber \]

First, consider the SO42- ions. There is one on the left side of the equation, and three on the right side. Add a coefficient of three to the left side.	\[Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + Mn(s) \nonumber \]
Next, check the Mn atoms. There is one on the right side, but now there are three on the left side from the previous adjustment. Add a coefficient of three on the right side.	\[Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + 3Mn(s)\nonumber \]
Consider Al. There is one atom on the left side and two on the right side. Add a coefficient of two on the left side. Make sure there are equal numbers of each atom on each side.	\[2Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + 3 Mn(s)\nonumber \]

Example \(\PageIndex{3}\):

\[P_4S_3 + KClO_3 \rightleftharpoons P_2O_5 + KCl + SO_2 \nonumber \]

This problem is more difficult. First, look at the P atoms. There are four on the reactant side and two on the product side. Add a coefficient of two to the product side.	\[P_4S_3 + KClO_3 \rightleftharpoons 2P_2O_5 + KCl + SO_2\nonumber \]
Next, consider the sulfur atoms. There are three on the left and one on the right. Add a coefficient of three to the right side.	\[P_4S_3 + KClO_3 \rightleftharpoons 2P_2O_5 + KCl + 3SO_2\nonumber \]
Now look at the oxygen atoms. There are three on the left and 16 on the right. Adding a coefficient of 16 to the KClO3 on the left and the KCl on the right preserves equal numbers of K and Cl atoms, but increases the oxygen.	\[P_4S_3 + 16KClO_3 \rightleftharpoons 2P_2O_5 + 16KCl + 3 SO_2\nonumber \]
Tripling the other three species (P4S3, P2O5, and SO2) balances the rest of the atoms.	\[3P_4S_3 + 16 KClO_3 \rightleftharpoons 2(3)P_2O_5 + 16KCl + 3(3)SO_2\nonumber \]
Simplify and check.	\[3P_4S_3 + 16KClO_3 \rightleftharpoons 6P_2O_5 + 16KCl + 9SO_2\nonumber \]

Balanced chemical equations can now be applied to the concept of chemical equilibrium, the state in which the reactants and products experience no net change over time. This occurs when the forward and reverse reactions occur at equal rates. The equilibrium constant is used to determine the amount of each compound that present at equilibrium. Consider a chemical reaction of the following form:

\[ aA + bB \rightleftharpoons cC + dD\nonumber \]

For this equation, the equilibrium constant is defined as:

\[ K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b} \nonumber \]

The activities of the products are in the numerator, and those of the reactants are in the denominator. For Kc, the activities are defined as the molar concentrations of the reactants and products ([A], [B] etc.). The lower case letters are the stoichiometric coefficients that balance the equation.

An important aspect of this equation is that pure liquids and solids are not included. This is because their activities are defined as one, so plugging them into the equation has no impact. This is due to the fact that pure liquids and solids have no effect on the physical equilibrium; no matter how much is added, the system can only dissolve as much as the solubility allows. For example, if more sugar is added to a solution after the equilibrium has been reached, the extra sugar will not dissolve (assuming the solution is not heated, which would increase the solubility). Because adding more does not change the equilibrium, it is not accounted for in the expression.

The following are concepts that apply when adjusting K in response to changes to the corresponding balanced equation:

When the equation is reversed, the value of K is inverted.
When the coefficients in a balanced equation are multiplied by a common factor, the equilibrium constant is raised to the power of the corresponding factor.
When the coefficients in a balanced equation are divided by a common factor, the corresponding root of the equilibrium constant is taken.
When individual equations are combined, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.

A balanced equation is very important in using the constant because the coefficients become the powers of the concentrations of products and reactants. If the equation is not balanced, then the constant is incorrect.

For gas-phase equilibria, the equation is a function of the reactants' and products' partial pressures. The equilibrium constant is expressed as follows:

\[ K_p = \dfrac{P_C^c P_D^d}{P_A^a P_B^b} \nonumber \]

P represents partial pressure, usually in atmospheres. As before, pure solids and liquids are not accounted for in the equation. Kc and Kp are related by the following equation:

\[ K_p = K_c(RT)^{\Delta n} \nonumber \]

where

\[ \Delta n = (c+d) - (a+b) \nonumber \]

This represents the change in gas molecules. a,b,c and d are the stoichiometric coefficients of the gas molecules found in the balanced equation.

Neither Kc nor Kp have units. This is due to their formal definitions in terms of activities. Their units cancel in the calculation, preventing problems with units in further calculations.

\[ PbI_2 \rightleftharpoons Pb \; (aq) + I \; (aq) \nonumber \]

First, balance the equation.

Check the Pb atoms. There is one on each side, so lead can be left alone for now. Next check the I atoms. There are two on the left side and one on the right side. To fix this, add a coefficient of two to the right side.	\[PbI_2 \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\nonumber \]
Check to make ensure the numbers are equal.	\[PbI_2 \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\nonumber \]

Next, calculate find Kc. Use these concentrations: Pb- 0.3 mol/L, I- 0.2 mol/L, PbI2- 0.5 mol/L

\[ K_c = \dfrac{(0.3) * (0.2)^2}{(0.5)} \nonumber \]

\[K_c= 0.024\nonumber \]

Note: If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I- would not have been squared. This would have given an incorrect answer.

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Example \(\PageIndex{5}\)

\[SO_2 \; (g) + O_2 \; (g) \rightleftharpoons SO_3 \; (g) \nonumber \]

First, make sure the equation is balanced.

Check to make sure S is equal on both sides. There is one on each side. Next look at the O. There are four on the left side and three on the right. Adding a coefficient to the O2 on the left is ineffective, as the S on right must also be increased. Instead, add a coefficient to the SO2 on the left and the SO3 on the right.	\[2SO_2 + O_2 \rightleftharpoons 2SO_3\nonumber \]
The equation is now balanced.	\[2SO_2 + O_2 \rightleftharpoons 2SO_3\nonumber \]

Calculate Kp. The partial pressures are as follows: SO2- 0.25 atm, O2- 0.45 atm, SO3- 0.3 atm

\( K_p = \dfrac{(0.3)^2}{(0.25)^2 \times (0.45)} \)

\( K_p= 3.2\)