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Balancing Chemical Equations: Practice and Review

Balancing Chemical Equations: Practice and Review

Of all the skills to know for 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 reactions 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 balancing that first. Then, move on to the atom that shows up the second least number of times, and so on. At the end, make sure to count the number of atoms of each element on each side again, just to be sure.

Let’s illustrate this with an example:

P 4 O 10 + H 2 O → H 3 PO 4

First, let’s look at the element that appears least often. Notice that oxygen occurs twice on the left hand side, so that is not a good element to start out with. We could either start with phosphorus or hydrogen, so let’s start with phosphorus. There are four atoms of phosphorus on the left hand side, but only one on the right hand side. So, we can put the coefficient of 4 on the molecule that has phosphorous on the right hand side to balance them out.

P 4 O 10 + H 2 O → 4 H 3 PO 4

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 H 3 PO 4 , and we have four molecules). So, to balance those out, we have to put a six in front of H 2 O on the left.

P 4 O 10 + 6 H 2 O → 4 H 3 PO 4

At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P 4 O 10 and six from H 2 O 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

Balancing Chemical Equations Practice Problems

Try to balance these ten equations on your own, then check the answers below. They range in difficulty level, so don’t get discouraged if some of them seem too hard. Just remember to start with the element that shows up the least, and proceed from there. The best way to approach these problems is slowly and systematically. Looking at everything at once can easily get overwhelming. Good luck!

Complete Solutions:

1. co 2 + h 2 o → c 6 h 12 o 6 + o 2.

The first step 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 on the carbon-containing molecule on the left.

6CO 2 + H 2 O → C 6 H 12 O 6 + O 2

Next, let’s look at hydrogen. There are two hydrogen atoms on the left and twelve on the right. So, we will add a coefficient of six on the hydrogen-containing molecule on the left.

6CO 2 + 6H 2 O → C 6 H 12 O 6 + O 2

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). On the right, there are eight oxygen molecules. Now, we have two options to even out the right hand side: We can either multiply C 6 H 12 O 6 or O 2 by a coefficient. However, if we change C 6 H 12 O 6 , 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 O 2 . So, we can add a coefficient of six to the O 2 on the right. Our final answer will be:

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2

2. SiCl 4 + H 2 O → H 4 SiO 4 + HCl

The only element that occurs more than once on the same side of the equation here is hydrogen, so we can start with any other element. Let’s start by looking at silicon. Notice that there is only one atom of silicon on either side, so we do not need to add any coefficients yet. Next, let’s look at chlorine. There are four chlorine atoms on the left side and only one on the right. So, we will add a coefficient of four on the right.

SiCl 4 + H 2 O → H 4 SiO 4 + 4HCl

Next, let’s look at oxygen. Remember that we first want to analyze all the elements that only occur once on one side of the equation. There is only one oxygen atom on the left, but four on the right. So, we will add a coefficient of four on the left hand side of the equation.

SiCl 4 + 4H 2 O → H 4 SiO 4 + 4HCl

We are almost done! Now, we just have to check the number of hydrogen atoms on each side. The left has eight and the right also has eight, so we are done. Our final answer is

As always, make sure to double check that the number of atoms of each element balances on each side before continuing.

3. Al + HCl → AlCl 3 + H 2

This problem is a bit tricky, so be careful. Whenever a single atom is alone on either side of the equation, it is easiest to start with that element. So, we will start by counting the aluminum atoms on both sides. There is one on the left and one on the right, so we do not need to add any coefficients yet. Next, let’s look at hydrogen. There is also one on the left, but two on the right. So, we will add a coefficient of two on the left.

Al + 2HCl → AlCl 3 + H 2

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 AlCl 3 by two:

Al + 6HCl → 2AlCl 3 + H 2

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.

2Al + 6HCl → 2AlCl 3 + H 2

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:

2Al + 6HCl → 2AlCl 3 + 3H 2

4. Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2

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.

Na 2 CO 3 + HCl → 2NaCl + H 2 O + CO 2

Next, let’s look at carbon. There is one on the left and one on the right, so there are no coefficients to add. Since oxygen occurs in more than one place on the left, we will save it for last. Instead, look at hydrogen. There is one on the left and two on the right, so we will add a coefficient to the left.

Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2

Then, looking at chlorine, we see that it is already balanced with two on each side. Now we can go back to look at oxygen. There are three on the left and three on the right, so our final answer is

5. C 7 H 6 O 2 + O 2 → CO 2 + H 2 O

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.

C 7 H 6 O 2 + O 2 → 7CO 2 + H 2 O

Then, for hydrogen, there are six atoms on the left and two on the right. So, we will add a coefficient of three on the right.

C 7 H 6 O 2 + O 2 → 7CO 2 + 3H 2 O

Now, for oxygen, things will get a little tricky. Oxygen occurs in every molecule in the equation, so we have to be very careful when balancing it. There are four atoms of oxygen on the left and 17 on the right. There is no obvious way to balance these numbers, so we must use a little trick: fractions. Now, when writing our final answer, we cannot include fractions as it is not proper form, but it sometimes helps to use them to solve the problem. Also, try to avoid over-manipulating organic molecules. You can easily identify organic molecules, otherwise known as CHO molecules, because they are made up of only carbon, hydrogen, and oxygen. We don’t like to work with these molecules, because they are rather complex. Also, larger molecules tend to be more stable than smaller molecules, and less likely to react in large quantities.

So, to balance out the four and seventeen, we can multiply the O 2 on the left by 7.5. That will give us

C 7 H 6 O 2 + 7.5O 2 → 7CO 2 + 3H 2 O

Remember, fractions (and decimals) are not allowed in formal balanced equations, so multiply everything by two to get integer values. Our final answer is now

2C 7 H 6 O 2 + 15O 2 → 14CO 2 + 6H 2 O

6. Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3­-

We can start by balancing the iron on both sides. The left has two while the right only has one. So, we will add a coefficient of two to the right.

Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + 2Fe(OH) 3­-

Then, we can look at sulfur. There are three on the left, but only one on the right. So, we will add a coefficient of three to the right hand side.

Fe 2 (SO 4 ) 3 + KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

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

Fe 2 (SO 4 ) 3 + 6KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

7. Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3

Looking at calcium, we see that there are three on the left and one on the right, so we can add a coefficient of three on the right to balance them out.

Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Then, for phosphorus, we see that there are two on the left and four on the right. To balance these, add a coefficient of two on the left.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

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.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 6CaSiO 3

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.

2Ca 3 (PO 4 ) 2 + 6SiO 2 → P 4 O 10 + 6CaSiO 3

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

8. KClO 3 → KClO 4 + KCl

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.

4KClO 3 → 3KClO 4 + KCl

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

9. Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4

We can start here by balancing the aluminum atoms on both sides. The left has two molecules while the right only has one, so we will add a coefficient of two on the right.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + CaSO 4

Now, we can check sulfur. There are three on the left and only one on the right, so adding a coefficient of three will balance these.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Moving right along to calcium, there is only one on the left but three on the right, so we should add a coefficient of three.

Al 2 (SO 4 ) 3 + 3Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Double-checking all the atoms, we see that all the elements are balanced, so our final equation is

10. H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

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.

H 2 SO 4 + HI → H 2 S + I 2 + 4H 2 O

There is only one iodine on the left and two on the right, so a simple coefficient change can balance those.

H 2 SO 4 + 2HI → H 2 S + I 2 + 4H 2 O

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 H 2 SO 4 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 H 2 SO 4 already has two. So, we will change the coefficient from 2 to 8.

H 2 SO 4 + 8HI → H 2 S + I 2 + 4H 2 O

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

H 2 SO 4 + 8HI → H 2 S + 4I 2 + 4H 2 O

As with most skills, practice makes perfect when learning how to balance chemical equations. Keep working hard and try to do as many problems as you can to help you hone your balancing skills.

Do you have any tips or tricks to help you balance chemical equations? Let us know in the comments!

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Chemistry library

Unit 5: lesson 1.

Balancing chemical equations

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2FeCl 3 + MgO ---> Fe 2 O 3 + 3MgCl 2 The Fe also gets balanced in this step.
2FeCl 3 + 3MgO ---> Fe 2 O 3 + 3MgCl 2 The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
3Li + H 3 PO 4 ---> H 2 + Li 3 PO 4
3Li + 2H 3 PO 4 ---> 3H 2 + Li 3 PO 4 Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
6Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
ZnS + (3/2)O 2 ---> ZnO + SO 2
2ZnS + 3O 2 ---> 2ZnO + 2SO 2
FeS 2 + (5/2)Cl 2 ---> FeCl 3 + S 2 Cl 2
2FeS 2 + 5Cl 2 ---> 2FeCl 3 + 2S 2 Cl 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + H 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + (3/2)H 2
2Fe + 6HC 2 H 3 O 2 ---> 2Fe(C 2 H 3 O 2 ) 3 + 3H 2
H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
2H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + 2H 2 O(ℓ)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + CO 2 (g)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
Fe 2 O 3 (s) + 3 ⁄ 2 C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
2Fe 2 O 3 (s) + 3C(s) ---> 4Fe(s) + 3CO 2 (g)
2C 5 H 11 NH 2 + O 2 ---> CO 2 + 13H 2 O + NO 2
2C 5 H 11 NH 2 + O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
2C 5 H 11 NH 2 + 37 ⁄ 2 O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
4, 37 ---> 20, 26, 4
CO 2 + 3 ⁄ 8 S 8 ---> CS 2 + SO 2 Most of the time the fraction used to balance is something with a 2 in the denominator: 1 ⁄ 2 or 5 ⁄ 2 or 13 ⁄ 2 , for example. Not too often does one see 3 ⁄ 8 . Pretty tricky!
8CO 2 + 3S 8 ---> 8CS 2 + 8SO 2
P 4 + 3O 2 ---> 2P 2 O 3 This depends on seeing that the oxygen on the left comes in twos and the oxygen on the right comes in threes. So, you use a three and a two to arrive at six oxygens on each side. Least common multiple, baby!!
P 4 + O 2 ---> 2P 2 O 3 Then, the oxygen gets balanced: P 4 + 3O 2 ---> 2P 2 O 3
1 ⁄ 2 P 4 + O 2 ---> P 2 O 3 Let us balance the oxygen with a fraction as well: 1 ⁄ 2 P 4 + 3 ⁄ 2 O 2 ---> P 2 O 3 Finally, multiply through by two: P 4 + 3O 2 ---> 2P 2 O 3

How to Balance Chemical Equations

Last Updated: January 31, 2023 References

This article was co-authored by Bess Ruff, MA . Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 4,620,859 times.

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.

Doing a Traditional Balance

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Completing an Algebraic Balance

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

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Write a Chemical Equation

About This Article

Bess Ruff, MA

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! Did this summary help you? Yes No

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Chemistry LibreTexts

5.1.1: Practice Problems- Writing and Balancing Chemical Equations

PROBLEM \(\PageIndex{1}\)

Balance the following equations:

\(\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{2HCl}(aq)\)

\(\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow \ce{3Cu(NO3)2}(aq)+\ce{4H2O}(l)+\ce{2NO}(g)\)

\(\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{2HI}(s)\)

\(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

\(\ce{2Na}(s)+\ce{2H2O}(l)\rightarrow \ce{2NaOH}(aq)+\ce{H2}(g)\)

\(\ce{(NH4)2Cr52O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{4H2O}(g)\)

\(\ce{P4}(s)+\ce{6Cl2}(g)\rightarrow \ce{4PCl3}(l)\)

\(\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{2Cl2}(g)\)

PROBLEM \(\PageIndex{2}\)

\(\ce{4Ag}(s)+\ce{2H2S}(g)+\ce{O2}(g)\rightarrow \ce{2Ag2S}(s)+\ce{2H2O}(l)\)

\(\ce{P4}(s)+\ce{5O2}(g)\rightarrow \ce{P4O10}(s)\)

\(\ce{2Pb}(s)+\ce{2H2O}(l)+\ce{O2}(g)\rightarrow \ce{2Pb(OH)2}(s)\)

\(\ce{3Fe}(s)+\ce{4H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{4H2}(g)\)

\(\ce{Sc2O3}(s)+\ce{3SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)\)

\(\ce{Ca3(PO4)2}(aq)+\ce{4H3PO4}(aq)\rightarrow \ce{3Ca(H2PO4)2}(aq)\)

\(\ce{2Al}(s)+\ce{3H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{3H2}(g)\)

\(\ce{TiCl4}(s)+\ce{2H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{4HCl}(g)\)

PROBLEM \(\PageIndex{3}\)

Write a balanced molecular equation describing each of the following chemical reactions.

\(\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(g)\)

\(\ce{2C4H10}(g)+\ce{13O2}(g)\rightarrow \ce{8CO2}(g)+\ce{10H2O}(g)\)

\(\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)\)

\(\ce{2H2O}(g)+\ce{2Na}(s)\rightarrow \ce{2NaOH}(s)+\ce{H2}(g)\)

PROBLEM \(\PageIndex{4}\)

Write a balanced equation describing each of the following chemical reactions.

\(\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)\)

\(\ce{2Al}(s)+\ce{3I2}(s)\rightarrow \ce{Al2I6}(s)\)

\(\ce{2NaCl}(s)+\ce{H2SO4}(aq)\rightarrow \ce{2HCl}(g)+\ce{Na2SO4}(aq)\)

\(\ce{H3PO4}(aq)+\ce{KOH}(aq)\rightarrow \ce{KH2PO4}(aq)+\ce{H2O}(l)\)

PROBLEM \(\PageIndex{5}\)

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.

Ba(NO 3 ) 2 , KClO 3

\(\ce{2Ba(NO3)2}(s)\rightarrow \ce{2BaO}(s)+\ce{2N2}(g)+\ce{5O2}(g)\)

\(\ce{2Mg}(s)+\ce{O2}(g)\rightarrow \ce{2MgO}(s)\) ; \(\ce{4Al}(s)+\ce{3O2}(g)\rightarrow \ce{2Al2O3}(g)\); \(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

PROBLEM \(\PageIndex{6}\)

Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).

\(\ce{4HF}(aq)+\ce{SiO2}(s)\rightarrow \ce{SiF4}(g)+\ce{2H2O}(l)\)

\(\ce{CaCl2}(aq)+\ce{2NaF}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{CaF2}(s)\)

PROBLEM \(\PageIndex{7}\)

A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.

\(\ce{CaO}(s)+\ce{H2O}(l)\rightarrow \ce{Ca(OH)2}(s)\)

\(\ce{Ca(OH)2}(s)+\ce{MgCl2}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{CaCl2}(aq)\)

\(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{2H2O}(l)\)

\(\ce{MgCl2}(s)\rightarrow \ce{Mg}(s)+\ce{Cl2}(g)\)

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).

Balancing Equations Test Questions

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Chemical reactions have the same number of atoms before the reaction as after the reaction. Balancing chemical equations is a basic skill in chemistry and testing yourself helps retain important information. This collection of ten chemistry test questions will give you practice in how to balance chemical reactions .

Balance the following equation:

__ SnO 2 + __ H 2 → __ Sn + __ H 2 O

__ KOH + __ H 3 PO 4 → __ K 3 PO 4 + __ H 2 O

__ KNO 3 + __ H 2 CO 3 → __ K 2 CO 3 + __ HNO 3

Balance the following equation: __ Na 3 PO 4 + __ HCl → __ NaCl + __ H 3 PO 4

__ TiCl 4 + __ H 2 O → __ TiO 2 + __ HCl

__ C 2 H 6 O + __ O 2 → __ CO 2 + __ H 2 O

__ Fe + __ HC 2 H 3 O 2 → __ Fe(C 2 H 3 O 2 ) 3 + __ H 2

__ NH 3 + __ O 2 → __ NO + __ H 2 O

__ B 2 Br 6 + __ HNO 3 → __ B(NO 3 ) 3 + __ HBr

Question 10

__ NH 4 OH + __ Kal(SO 4 ) 2 ·12H 2 O → __ Al(OH) 3 + __ (NH 4 ) 2 SO 4 + __ KOH + __ H 2 O

1. 1 SnO 2 + 2 H 2 → 1 Sn + 2 H 2 O 2. 3 KOH + 1 H 3 PO 4 → 1 K 3 PO 4 + 3 H 2 O 3. 2 KNO 3 + 1 H 2 CO 3 → 1 K 2 CO 3 + 2 HNO 3 4. 1 Na 3 PO 4 + 3 HCl → 3 NaCl + 1 H 3 PO 4 5. 1 TiCl 4 + 2 H 2 O → 1 TiO 2 + 4 HCl 6. 1 C 2 H 6 O + 3 O 2 → 2 CO 2 + 3 H 2 O 7. 2 Fe + 6 HC 2 H 3 O 2 → 2 Fe(C 2 H 3 O 2 ) 3 + 3 H 2 8. 4 NH 3 + 5 O 2 → 4 NO + 6 H 2 O 9. 1 B 2 Br 6 + 6 HNO 3 → 2 B(NO 3 ) 3 + 6 HBr 10. 4 NH 4 OH + 1 Kal(SO 4 ) 2 ·12H 2 O → 1 Al(OH) 3 + 2 (NH 4 ) 2 SO 4 + 1 KOH + 12 H 2 O

Tips for Balancing Equations

When balancing equations, remember chemical reactions must satisfy conservation of mass. Check your work to make certain you have the same number and type of atoms on the reactants side as on the products side. A coefficient (number in front of a chemical) is multiplied by all the atoms in that chemical. A subscript (lower number) is only multiplied by the number of atoms it immediately follows. If there is no coefficient or subscript, that is the same as a number "1" (which is not written in chemical formulas).

how to solve chemical equations problems

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