20 Balanced Chemical Equations With Answers

[Martha Vanschaick]

Howeverthis equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.

This method uses algebraic equations to find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, y, z), and a series of equations are set up based on the number of each type of atom.

Of all the skills to know about in chemistry, balancing chemical equations is perhaps the most important to master. So many parts of chemistry depend on this vital skill, including stoichiometry, reaction analysis, and lab work. This comprehensive guide will show you the steps to balance even the most challenging reactions and will walk you through a series of examples, from simple to complex.

The ultimate goal for balancing chemical equations is to make both sides of the reaction, the reactants and the products, equal in the number of atoms per element. This stems from the universal law of the conservation of mass, which states that matter can neither be created nor destroyed. So, if we start with ten atoms of oxygen before a reaction, we need to end up with ten atoms of oxygen after a reaction. This means that chemical reactions do not change the actual building blocks of matter; rather, they just change the arrangement of the blocks. An easy way to understand this is to picture a house made of blocks. We can break the house apart and build an airplane, but the color and shape of the actual blocks do not change.

But how do we go about balancing these equations? We know that the number of atoms of each element needs to be the same on both sides of the equation, so it is just a matter of finding the correct coefficients (numbers in front of each molecule) to make that happen. It is best to start with the atom that shows up the least number of times on one side, and balance that first. Then, move on to the atom that shows up the second least number of times, and so on. In the end, make sure to count the number of atoms of each element on each side again, just to be sure.

Now we can check hydrogen. We still want to avoid balancing oxygen, because it occurs in more than one molecule on the left-hand side. It is easiest to start with molecules that only appear once on each side. So, there are two molecules of hydrogen on the left-hand side and twelve on the right-hand side (notice that there are three per molecule of H3PO4, and we have four molecules). So, to balance those out, we have to put a six in front of H2O on the left.

At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P4O10 and six from H2O for a total of 16. On the right, we have 16 as well (four per molecule, with four molecules). So, oxygen is already balanced. This gives us the final balanced equation of

The first step to balancing chemical equations is to focus on elements that only appear once on each side of the equation. Here, both carbon and hydrogen fit this requirement. So, we will start with carbon. There is only one atom of carbon on the left-hand side, but six on the right-hand side. So, we add a coefficient of six to the carbon-containing molecule on the left.

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (62 + 61). On the right, there are eight oxygen molecules. Now, we have two options to even out the right-hand side: We can either multiply C6H12O6 or O2 by a coefficient. However, if we change C6H12O6, the coefficients for everything else on the left-hand side will also have to change, because we will be changing the number of carbon and hydrogen atoms. To prevent this, it usually helps to only change the molecule containing the fewest elements; in this case, the O2. So, we can add a coefficient of six to the O2 on the right. Our final answer will be:

Next, we will look at chlorine. There are now two on the left, but three on the right. Now, this is not as straightforward as just adding a coefficient to one side. We need the number of chlorine atoms to be equal on both sides, so we need to get two and three to be equal. We can accomplish this by finding the lowest common multiple. In this case, we can multiply two by three and three by two to get the lowest common multiple of six. So, we will multiply 2HCl by three and AlCl3 by two:

We have looked at all the elements, so it is easy to say that we are done. However, always make sure to double-check. In this case, because we added a coefficient to the aluminum-containing molecule on the right-hand side, aluminum is no longer balanced. There is one on the left but two on the right. So, we will add one more coefficient.

We are not quite done yet. Looking over the equation one final time, we see that hydrogen has also been unbalanced. There are six on the left but two on the right. So, with one final adjustment, we get our final answer:

Hopefully, by this point, balancing equations is becoming easier and you are getting the hang of it. Looking at sodium, we see that it occurs twice on the left, but once on the right. So, we can add our first coefficient to the NaCl on the right.

We can start balancing this equation by looking at either carbon or hydrogen. Looking at carbon, we see that there are seven atoms on the left and only one on the right. So, we can add a coefficient of seven on the right.

We are almost done. All that is left is to balance the potassium. There is one atom on the left and six on the right, so we can balance these by adding a coefficient of six. Our final answer, then, is

Notice that by doing so, we changed the number of calcium atoms on the left. Every time you add a coefficient, double check to see if the step affects any elements you have already balanced. In this case, the number of calcium atoms on the left has increased to six while it is still three on the right, so we can change the coefficient on the right to reflect this change.

Since oxygen occurs in every molecule in the equation, we will skip it for now. Focusing on silicon, we see that there is one on the left, but six on the right, so we can add a coefficient to the left.

Now, we will check the number of oxygen atoms on each side. The left has 28 atoms and the right also has 28. So, after checking that all the other atoms are the same on both sides as well, we get a final answer of

This problem is particularly tricky because every atom, except oxygen, occurs in every molecule in the equation. So, since oxygen appears the least number of times, we will start there. There are three on the left and four on the right. To balance these, we find the lowest common multiple; in this case, 12. By adding a coefficient of four on the left and three on the right, we can balance the oxygens.

Now, we can check potassium and chlorine. There are four potassium molecules on the left and four on the right, so they are balanced. Chlorine is also balanced, with four on each side, so we are finished, with a final answer of

Since hydrogen occurs more than once on the left, we will temporarily skip it and move to sulfur. There is one atom on the left and one on the right, so there is nothing to balance yet. Looking at oxygen, there are four on the left and one on the right, so we can add a coefficient of four to balance them.

Now, we can look at the most challenging element: hydrogen. On the left, there are four and on the right, there are ten. So, we know we have to change the coefficient of either H2SO4 or HI. We want to change something that will require the least amount of tweaking afterwards, so we will change the coefficient of HI. To get the left-hand side to have ten atoms of hydrogen, we need HI to have eight atoms of hydrogen, since H2SO4 already has two. So, we will change the coefficient from 2 to 8.

However, this also changes the balance for iodine. There are now eight on the left, but only two on the right. To fix this, we will add a coefficient of 4 on the right. After checking that everything else balances out as well, we get a final answer of

Chemical equations are symbolic representations of chemical reactions that express the reactants and products in terms of their respective chemical formulae. They also use symbols to represent factors such as reaction direction and the physical states of the reacting entities.

Answer. The chemical equation must be balanced in order to obey the law of conservation of mass. A chemical equation is said to be balanced when the number of different atoms of elements in the reactants side equals the number of atoms in the products side. Balancing chemical equations is a trial-and-error process.

Q11. What is meant by the skeletal type chemical equation? What does it represent? Using the equation for electrolytic decomposition of water, differentiate between a skeletal chemical equation and a balanced chemical equation.

Answer. The primary distinction between a balanced equation and a skeleton equation is that the balanced equation provides the actual number of molecules of each reactant and product involved in the chemical reaction, whereas a skeleton equation only provides the reactants. Furthermore, a balanced equation may or may not contain stoichiometric coefficients, whereas a skeleton equation does not.

Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.

Suppose we know we have 123.2 g of Cl2. How can we determine how many moles of AlCl3 we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl2 to an amount of AlCl3, we need to convert the given amount of Cl2 into moles. We know how to do this by simply using the molar mass of Cl2 as a conversion factor. The molar mass of Cl2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:

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