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Introduction

As we have already mentioned, a number of changes can occur when elements react with one another. These changes may either be physical or chemical. One way of representing these changes is through balanced chemical equations. A chemical equation describes a chemical reaction by using symbols for the elements involved. For example, if we look at the reaction between iron (FeFe) and sulphur (SS) to form iron sulphide (FeSFeS), we could represent these changes either in words or using chemical symbols:

iron+sulphuriron sulphideiron+sulphuriron sulphide

or

Fe+SFeSFe+SFeS

Another example would be:

ammonia+oxygennitric oxide+waterammonia+oxygennitric oxide+water

or

4NH3+5O2 4NO+6H2O4NH3+5O24NO+6H2O

Compounds on the left of the arrow are called the reactants and these are needed for the reaction to take place. In this equation, the reactants are ammonia and oxygen. The compounds on the right are called the products and these are what is formed from the reaction.

In order to be able to write a balanced chemical equation, there are a number of important things that need to be done:

  1. Know the chemical symbols for the elements involved in the reaction
  2. Be able to write the chemical formulae for different reactants and products
  3. Balance chemical equations by understanding the laws that govern chemical change
  4. Know the state symbols for the equation

We will look at each of these steps separately in the next sections.

Chemical symbols

It is very important to know the chemical symbols for common elements in the Periodic Table, so that you are able to write chemical equations and to recognise different compounds.

Revising common chemical symbols

  • Write down the chemical symbols and names of all the elements that you know.
  • Compare your list with another learner and add any symbols and names that you don't have.
  • Spend some time, either in class or at home, learning the symbols for at least the first twenty elements in the periodic table. You should also learn the symbols for other common elements that are not in the first twenty.
  • Write a short test for someone else in the class and then exchange tests with them so that you each have the chance to answer one.

Writing chemical formulae

A chemical formula is a concise way of giving information about the atoms that make up a particular chemical compound. A chemical formula shows each element by its symbol and also shows how many atoms of each element are found in that compound. The number of atoms (if greater than one) is shown as a subscript.

Examples:

CH4CH4 (methane)

Number of atoms: (1×carbon)+(4×hydrogen)=5(1×carbon)+(4×hydrogen)=5 atoms in one methane molecule

H2SO4H2SO4 (sulphuric acid)

Number of atoms: (2×hydrogen)+(1×sulphur)+(4×oxygen)=7(2×hydrogen)+(1×sulphur)+(4×oxygen)=7 atoms in one molecule of sulphuric acid

A chemical formula may also give information about how the atoms are arranged in a molecule if it is written in a particular way. A molecule of ethane, for example, has the chemical formula C2H6C2H6. This formula tells us how many atoms of each element are in the molecule, but doesn't tell us anything about how these atoms are arranged. In fact, each carbon atom in the ethane molecule is bonded to three hydrogen atoms. Another way of writing the formula for ethane is CH3CH3CH3CH3. The number of atoms of each element has not changed, but this formula gives us more information about how the atoms are arranged in relation to each other.

The slightly tricky part of writing chemical formulae comes when you have to work out the ratio in which the elements combine. For example, you may know that sodium (NaNa) and chlorine (ClCl) react to form sodium chloride, but how do you know that in each molecule of sodium chloride there is only one atom of sodium for every one atom of chlorine? It all comes down to the valency of an atom or group of atoms. Valency is the number of bonds that an element can form with another element. Working out the chemical formulae of chemical compounds using their valency, will be covered in Grade 11. For now, we will use formulae that you already know.

Balancing chemical equations

The law of conservation of mass

In order to balance a chemical equation, it is important to understand the law of conservation of mass.

Definition 1: The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes acting inside the system. Matter can change form, but cannot be created or destroyed. For any chemical process in a closed system, the mass of the reactants must equal the mass of the products.

In a chemical equation then, the mass of the reactants must be equal to the mass of the products. In order to make sure that this is the case, the number of atoms of each element in the reactants must be equal to the number of atoms of those same elements in the products. Some examples are shown below:

Example 1:

Fe+SFeSFe+SFeS
(1)

Figure 1
Figure 1 (CG10C5_001.png)

Reactants

Atomic mass of reactants=55,8u+32,1u=87,9uAtomic mass of reactants=55,8u+32,1u=87,9u

Number of atoms of each element in the reactants: (1×Fe)(1×Fe) and (1×S)(1×S)

Products

Atomic mass of products=55,8u+32,1u=87,9uAtomic mass of products=55,8u+32,1u=87,9u

Number of atoms of each element in the products: (1×Fe)(1×Fe) and (1×S)(1×S)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the equation is balanced.

Example 2:

H2 + O2 H2OH2+O2H2O
(2)

Figure 2
Figure 2 (CG10C5_002.png)

Reactants

Atomic mass of reactants=(1+1)+(16+16)=34uAtomic mass of reactants=(1+1)+(16+16)=34u

Number of atoms of each element in the reactants: (2×H)(2×H) and (2×O)(2×O)

Product

Atomic mass of product=(1+1+16)=18uAtomic mass of product=(1+1+16)=18u

Number of atoms of each element in the product: (2×H)(2×H) and (1×O)(1×O)

Since the total atomic mass of the reactants and the products is not the same and since there are more oxygen atoms in the reactants than there are in the product, the equation is not balanced.

Example 3:

NaOH + HCl NaCl + H2O NaOH + HCl NaCl + H2O
(3)

Figure 3
Figure 3 (CG10C5_003.png)

Reactants

Atomic mass of reactants=(23+6+1)+(1+35,4)=76,4uAtomic mass of reactants=(23+6+1)+(1+35,4)=76,4u

Number of atoms of each element in the reactants: (1×Na)+(1×O)+(2×H)+(1×Cl)(1×Na)+(1×O)+(2×H)+(1×Cl)

Products

Atomic mass of products=(23+35,4)+(1+1+16)=76,4uAtomic mass of products=(23+35,4)+(1+1+16)=76,4u

Number of atoms of each element in the products: (1×Na)+(1×O)+(2×H)+(1×Cl)(1×Na)+(1×O)+(2×H)+(1×Cl)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the equation is balanced.

We now need to find a way to balance those equations that are not balanced so that the number of atoms of each element in the reactants is the same as that for the products. This can be done by changing the coefficients of the molecules until the atoms on each side of the arrow are balanced. You will see later that these coefficients tell us something about the mole ratio in which substances react. They also tell us about the volume relationship between gases in the reactants and products.

Tip:

Coefficients

Remember that if you put a number in front of a molecule, that number applies to the whole molecule. For example, if you write 2H2O2H2O, this means that there are 2 molecules of water. In other words, there are 4 hydrogen atoms and 2 oxygen atoms. If we write 3HCl3HCl, this means that there are 3 molecules of HClHCl. In other words there are 3 hydrogen atoms and 3 chlorine atoms in total. In the first example, 2 is the coefficient and in the second example, 3 is the coefficient.

Activity: Balancing chemical equations

You will need: coloured balls (or marbles), prestik, a sheet of paper and coloured pens.

We will try to balance the following equation:

Al + O 2 Al 2 O 3 Al + O 2 Al 2 O 3
(4)
Take 1 ball of one colour. This represents a molecule of AlAl. Take two balls of another colour and stick them together. This represents a molecule of O2O2. Place these molecules on your left. Now take two balls of one colour and three balls of another colour to form a molecule of Al 2 O 3 Al 2 O 3 . Place these molecules on your right. On a piece of paper draw coloured circles to represent the balls. Draw a line down the center of the paper to represent the molecules on the left and on the right.

Count the number of balls on the left and the number on the right. Do you have the same number of each colour on both sides? If not the equation is not balanced. How many balls will you have to add to each side to make the number of balls the same? How would you add these balls?

You should find that you need 4 balls of one colour for AlAl and 3 pairs of balls of another colour (i.e. 6 balls in total) for O2O2 on the left side. On the right side you should find that you need 2 clusters of balls for Al 2 O 3 Al 2 O 3 . We say that the balanced equation is:

4 Al + 3 O 2 2 Al 2 O 3 4 Al + 3 O 2 2 Al 2 O 3
(5)

Repeat this process for the following reactions:

  • CH 4 + 2 O 2 CO 2 + 2 H 2 O CH 4 + 2 O 2 CO 2 + 2 H 2 O
  • 2 H 2 + O 2 2 H 2 O 2 H 2 + O 2 2 H 2 O
  • Zn + 2 HCl ZnCl 2 + H 2 Zn + 2 HCl ZnCl 2 + H 2

Steps to balance a chemical equation

When balancing a chemical equation, there are a number of steps that need to be followed.

  • STEP 1: Identify the reactants and the products in the reaction and write their chemical formulae.
  • STEP 2: Write the equation by putting the reactants on the left of the arrow and the products on the right.
  • STEP 3: Count the number of atoms of each element in the reactants and the number of atoms of each element in the products.
  • STEP 4: If the equation is not balanced, change the coefficients of the molecules until the number of atoms of each element on either side of the equation balance.
  • STEP 5: Check that the atoms are in fact balanced.
  • STEP 6 (we will look at this a little later): Add any extra details to the equation e.g. phase.

Exercise 1: Balancing chemical equations 1

Balance the following equation:

Mg + HCl MgCl2+ H2 Mg + HCl MgCl2+ H2
(6)

Solution
  1. Step 1. Because the equation has been written for you, you can move straight on to counting the number of atoms of each element in the reactants and products :

    Reactants: Mg=1atomMg=1atom; H=1atomH=1atom and Cl=1atomCl=1atom

    Products: Mg=1atomMg=1atom; H=2atomsH=2atoms and Cl=2atomsCl=2atoms

  2. Step 2. Balance the equation :

    The equation is not balanced since there are 2 chlorine atoms in the product and only 1 in the reactants. If we add a coefficient of 2 to the HClHCl to increase the number of HH and ClCl atoms in the reactants, the equation will look like this:

    Mg +2 HCl MgCl2+ H2 Mg +2 HCl MgCl2+ H2
    (7)

  3. Step 3. Check that the atoms are balanced :

    If we count the atoms on each side of the equation, we find the following:

    Reactants: Mg=1atomMg=1atom; H=2atomH=2atom and Cl=2atomCl=2atom

    Products: Mg=1atomMg=1atom; H=2atomH=2atom and Cl=2atomCl=2atom

    The equation is balanced. The final equation is:

    Mg +2 HCl MgCl2+ H2 Mg +2 HCl MgCl2+ H2
    (8)

Exercise 2: Balancing chemical equations 2

Balance the following equation:

CH 4 + O 2 CO 2 +H2O CH 4 + O 2 CO 2 +H2O
(9)

Solution
  1. Step 1. Count the number of atoms of each element in the reactants and products :

    Reactants: C=1C=1; H=4H=4 and O=2O=2

    Products: C=1C=1; H=2H=2 and O=3O=3

  2. Step 2. Balance the equation :

    If we add a coefficient of 2 to H2OH2O, then the number of hydrogen atoms in the reactants will be 4, which is the same as for the reactants. The equation will be:

    CH 4 + O 2 CO 2 +2H2O CH 4 + O 2 CO 2 +2H2O
    (10)

  3. Step 3. Check that the atoms balance :

    Reactants: C=1C=1; H=4H=4 and O=2O=2

    Products: C=1C=1; H=4H=4 and O=4O=4

    You will see that, although the number of hydrogen atoms now balances, there are more oxygen atoms in the products. You now need to repeat the previous step. If we put a coefficient of 2 in front of O2O2, then we will increase the number of oxygen atoms in the reactants by 2. The new equation is:

    CH 4 +2 O 2 CO 2 +2H2O CH 4 +2 O 2 CO 2 +2H2O
    (11)

    When we check the number of atoms again, we find that the number of atoms of each element in the reactants is the same as the number in the products. The equation is now balanced.

Exercise 3: Balancing chemical equations 3

Nitrogen gas reacts with hydrogen gas to form ammonia. Write a balanced chemical equation for this reaction.

Solution
  1. Step 1. Identify the reactants and the products and write their chemical formulae :

    The reactants are nitrogen (N2N2) and hydrogen (H2H2) and the product is ammonia (NH3NH3).

  2. Step 2. Write the equation so that the reactants are on the left and products on the right of the arrow :

    The equation is as follows:

    N 2 + H 2 NH 3 N 2 + H 2 NH 3
    (12)

  3. Step 3. Count the atoms of each element in the reactants and products :

    Reactants: N=2N=2 and H=2H=2

    Products: N=1N=1 and H=3H=3

  4. Step 4. Balance the equation :

    In order to balance the number of nitrogen atoms, we could rewrite the equation as:

    N 2 + H 2 2 NH 3 N 2 + H 2 2 NH 3
    (13)

  5. Step 5. Check that the atoms are balanced :

    In the above equation, the nitrogen atoms now balance, but the hydrogen atoms don't (there are 2 hydrogen atoms in the reactants and 6 in the product). If we put a coefficient of 3 in front of the hydrogen ( H 2 H 2 ), then the hydrogen atoms and the nitrogen atoms balance. The final equation is:

    N 2 +3 H 2 2 NH 3 N 2 +3 H 2 2 NH 3
    (14)

Exercise 4: Balancing chemical equations 4

In our bodies, sugar (C6H12O6C6H12O6) reacts with the oxygen we breathe in to produce carbon dioxide, water and energy. Write the balanced equation for this reaction.

Solution
  1. Step 1. Identify the reactants and products in the reaction and write their chemical formulae. :

    Reactants: sugar (C6H12O6C6H12O6) and oxygen (O2O2)

    Products: carbon dioxide (CO2CO2) and water (H2OH2O)

  2. Step 2. Write the equation by putting the reactants on the left of the arrow and the products on the right :
    C6H12O6+O2CO2+H2OC6H12O6+O2CO2+H2O
    (15)
  3. Step 3. Count the number of atoms of each element in the reactants and the number of atoms of each element in the products :

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=1C=1; H=2H=2 and O=3O=3

  4. Step 4. Change the coefficents of the molecules until the number of atoms of each element on either side of the equation balance. :

    It is easier to start with carbon as it only appears once on each side. If we add a 6 in front of CO2CO2, the equation looks like this:

    C6H12O6 + O2 6 CO2 + H2O C6H12O6+O26CO2+H2O
    (16)

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=6C=6; H=2H=2 and O=13O=13

  5. Step 5. Change the coefficients again to try to balance the equation. :

    Let's try to get the number of hydrogens the same this time.

    C6H12O6 + O2 6 CO2 +6 H2O C6H12O6+O26CO2+6H2O
    (17)

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=6C=6; H=12H=12 and O=18O=18

  6. Step 6. Now we just need to balance the oxygen atoms. :
    C6H12O6 +12 O2 6 CO2 +6 H2O C6H12O6+12O26CO2+6H2O
    (18)

    Reactants: C=6C=6; H=12H=12 and O=18O=18

    Products: C=6C=6; H=12H=12 and O=18O=18

This simulation allows you to practice balancing simple equations.

Figure 4
Figure 4 (balancing-chemical-equations-screenshot.png)
run demo

Balancing simple chemical equations

Balance the following equations:

  1. Hydrogen fuel cells are extremely important in the development of alternative energy sources. Many of these cells work by reacting hydrogen and oxygen gases together to form water, a reaction which also produces electricity. Balance the following equation: H2(g)+O2(g)H2O(l)H2(g)+O2(g)H2O(l)Click here for the solution
  2. The synthesis of ammonia (NH3NH3), made famous by the German chemist Fritz Haber in the early 20th century, is one of the most important reactions in the chemical industry. Balance the following equation used to produce ammonia: N2(g)+H2(g)NH3(g)N2(g)+H2(g)NH3(g) Click here for the solution
  3. Mg+P4Mg3P2Mg+P4Mg3P2 Click here for the solution
  4. Ca+H2OCa(OH)2+H2Ca+H2OCa(OH)2+H2Click here for the solution
  5. CuCO3+H2SO4CuSO4+H2O+ CO2 CuCO3+H2SO4CuSO4+H2O+ CO2 Click here for the solution
  6. CaCl2+Na2CO3CaCO3+NaCl CaCl2+Na2CO3CaCO3+NaClClick here for the solution
  7. C12H22O11+O2H2O+CO2C12H22O11+O2H2O+CO2
     
    Click here for the solution
  8. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid. Click here for the solution
  9. Ethane (C2H6C2H6) reacts with oxygen to form carbon dioxide and steam. Click here for the solution
  10. Ammonium carbonate is often used as a smelling salt. Balance the following reaction for the decomposition of ammonium carbonate: (NH4)2CO3(s)NH3(aq)+CO2(g)+H2O(l)(NH4)2CO3(s)NH3(aq)+CO2(g)+H2O(l)
     
    Click here for the solution

State symbols and other information

The state (phase) of the compounds can be expressed in the chemical equation. This is done by placing the correct label on the right hand side of the formula. There are only four labels that can be used:

  1. (g) for gaseous compounds
  2. (l) for liquids
  3. (s) for solid compounds
  4. (aq) for an aqueous (water) solution

Occasionally, a catalyst is added to the reaction. A catalyst is a substance that speeds up the reaction without undergoing any change to itself. In a chemical equation, this is shown by using the symbol of the catalyst above the arrow in the equation.

To show that heat is needed for a reaction, a Greek delta (ΔΔ) is placed above the arrow in the same way as the catalyst.

Tip:

You may remember from Physical and chemical change that energy cannot be created or destroyed during a chemical reaction but it may change form. In an exothermic reaction, ΔΔH is less than zero and in an endothermic reaction, ΔΔH is greater than zero. This value is often written at the end of a chemical equation.

Exercise 5: Balancing chemical equations 5

Solid zinc metal reacts with aqueous hydrochloric acid to form an aqueous solution of zinc chloride (ZnCl2ZnCl2)and hydrogen gas. Write a balanced equation for this reaction.

Solution

  1. Step 1. Identify the reactants and products and their chemical formulae :

    The reactants are zinc (ZnZn) and hydrochloric acid (HClHCl). The products are zinc chloride (ZnCl2ZnCl2) and hydrogen (H2H2).

  2. Step 2. Place the reactants on the left of the equation and the products on the right hand side of the arrow :
    Zn + HCl ZnCl2 + H2 Zn + HCl ZnCl2 + H2
    (19)
  3. Step 3. Balance the equation :

    You will notice that the zinc atoms balance but the chlorine and hydrogen atoms don't. Since there are two chlorine atoms on the right and only one on the left, we will give HClHCl a coefficient of 2 so that there will be two chlorine atoms on each side of the equation.

    Zn +2 HCl ZnCl2 + H2 Zn +2 HCl ZnCl2 + H2
    (20)

  4. Step 4. Check that all the atoms balance :

    When you look at the equation again, you will see that all the atoms are now balanced.

  5. Step 5. Ensure all details (e.g. state symbols) are added :

    In the initial description, you were told that zinc was a metal, hydrochloric acid and zinc chloride were in aqueous solutions and hydrogen was a gas.

    Zn(s) +2 HCl(aq) ZnCl2(aq) + H2(g) Zn(s) +2 HCl(aq) ZnCl2(aq) + H2(g)
    (21)

Exercise 6: Balancing chemical equations 5 (advanced)

Balance the following equation:

( NH4)2SO4 + NaOH NH3 + H2O + Na2SO4 ( NH4)2SO4 + NaOH NH3 + H2O + Na2SO4
(22)

In this example, the first two steps are not necessary because the reactants and products have already been given.

Solution

  1. Step 1. Balance the equation :

    With a complex equation, it is always best to start with atoms that appear only once on each side i.e. NaNa, NN and SS atoms. Since the SS atoms already balance, we will start with NaNa and NN atoms. There are two NaNa atoms on the right and one on the left. We will add a second NaNa atom by giving NaOHNaOH a coefficient of two. There are two NN atoms on the left and one on the right. To balance the NN atoms, NH3NH3 will be given a coefficient of two. The equation now looks as follows:

    ( NH4)2SO4 +2 NaOH 2 NH3 + H2O + Na2SO4 ( NH4)2SO4 +2 NaOH 2 NH3 + H2O + Na2SO4
    (23)

  2. Step 2. Check that all atoms balance :

    NN, NaNa and SS atoms balance, but OO and HH atoms do not. There are six OO atoms and ten HH atoms on the left, and five OO atoms and eight HH atoms on the right. We need to add one OO atom and two HH atoms on the right to balance the equation. This is done by adding another H2OH2O molecule on the right hand side. We now need to check the equation again:

    ( NH4)2SO4 +2 NaOH 2 NH3 +2 H2O + Na2SO4 ( NH4)2SO4 +2 NaOH 2 NH3 +2 H2O + Na2SO4
    (24)

    The equation is now balanced.

The following video explains some of the concepts of balancing chemical equations.

Figure 5
Khan Academy video on balancing chemical equations

Balancing more advanced chemical equations

Write balanced equations for each of the following reactions:

  1. Al2O3(s)+H2SO4(aq) Al2(SO4)3(aq)+3H2O(l)Al2O3(s)+H2SO4(aq)Al2(SO4)3(aq)+3H2O(l)
  2. Mg(OH)2(aq)+HNO3(aq) Mg(NO3)2(aq)+2H2O(l)Mg(OH)2(aq)+HNO3(aq)Mg(NO3)2(aq)+2H2O(l)
  3. Lead (II) nitrate solution reacts with potassium iodide solution.
  4. When heated, aluminium reacts with solid copper oxide to produce copper metal and aluminium oxide (Al2O3Al2O3).
  5. When calcium chloride solution is mixed with silver nitrate solution, a white precipitate (solid) of silver chloride appears. Calcium nitrate (Ca(NO3)2Ca(NO3)2) is also produced in the solution. Click here for the solution

Balanced equations are very important in chemistry. It is only by working with the balanced equations that chemists can perform many different calculations that tell them what quantity of something reacts. In a later chapter we will learn how to work with some of these calculations. We can interpret balanced chemical equations in terms of the conservation of matter, the conservation of mass or the conservation of energy.

Figure 6

Summary

  • A chemical equation uses symbols to describe a chemical reaction.
  • In a chemical equation, reactants are written on the left hand side of the equation and the products on the right. The arrow is used to show the direction of the reaction.
  • When representing chemical change, it is important to be able to write the chemical formula of a compound.
  • In any chemical reaction, the law of conservation of mass applies. This means that the total atomic mass of the reactants must be the same as the total atomic mass of the products. This also means that the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the product.
  • If the number of atoms of each element in the reactants is the same as the number of atoms of each element in the product, then the equation is balanced.
  • If the number of atoms of each element in the reactants is not the same as the number of atoms of each element in the product, then the equation is not balanced.
  • In order to balance an equation, coefficients can be placed in front of the reactants and products until the number of atoms of each element is the same on both sides of the equation.

End of chapter exercises

  1. Propane is a fuel that is commonly used as a heat source for engines and homes. Balance the following equation for the combustion of propane: C3 H8(l) + O2(g) CO2(g) + H2O(l) C3 H8(l) + O2(g) CO2(g) + H2O(l) Click here for the solution
  2. Aspartame, an artificial sweetener, has the formula C14H18N2O2C14H18N2O2. Write the balanced equation for its combustion (reaction with O2O2) to form CO2CO2 gas, liquid H2OH2O, and N2N2 gas. Click here for the solution
  3. Fe2(SO4)3+K(SCN) K3Fe(SCN)6+K2SO4Fe2(SO4)3+K(SCN)K3Fe(SCN)6+K2SO4 Click here for the solution
  4. Chemical weapons were banned by the Geneva Protocol in 1925. According to this protocol, all chemicals that release suffocating and poisonous gases are not to be used as weapons. White phosphorus, a very reactive allotrope of phosphorus, was recently used during a military attack. Phosphorus burns vigorously in oxygen. Many people got severe burns and some died as a result. The equation for this spontaneous reaction is: P4(s) + O 2 (g) P2O5(s) P4(s) + O 2 (g) P2O5(s)
    1. Balance the chemical equation.
    2. Prove that the law of conservation of mass is obeyed during this chemical reaction.
    3. Name the product formed during this reaction.
    4. Classify the reaction as endothermic or exothermic. Give a reason for your answer.
    5. Classify the reaction as a synthesis or decomposition reaction. Give a reason for your answer. Click here for the solution
    (DoE Exemplar Paper 2 2007)
  5. Mixing bleach (NaOClNaOCl) and ammonia (two common household cleaners) is very dangerous. When these two substances are mixed they produce toxic chloaramine (NH2ClNH2Cl) fumes. Balance the following equations that occur when bleach and ammonia are mixed:
    1. NaOCl(aq)+NH3(aq)NaONH3(aq)+Cl2(g)NaOCl(aq)+NH3(aq)NaONH3(aq)+Cl2(g)
    2. If there is more bleach than ammonia the following may occur: NaOCl+NH3NaOH+NCl3NaOCl+NH3NaOH+NCl3
      Nitrogen trichloride (NCl3NCl3) is highly explosive.
    3. If there is more ammonia than bleach the following may occur: NH3+NaOClNaOH+NH2ClNH3+NaOClNaOH+NH2Cl
      These two products then react with ammonia as follows:
      NH3+NH2Cl+NaOHN2H4+NaCl+H2ONH3+NH2Cl+NaOHN2H4+NaCl+H2O
      One last reaction occurs to stabilise the hydrazine and chloramine: NH2Cl+N2H4NH4Cl+N2NH2Cl+N2H4NH4Cl+N2
      This reaction is highly exothermic and will explode.
    Click here for the solution
  6. Balance the following chemical equation: N2O5NO2+O2N2O5NO2+O2
    Click here for the solution
  7. Sulphur can be produced by the Claus process. This two-step process involves reacting hydrogen sulphide with oxygen and then reacting the sulphur dioxide that is produced with more hydrogen sulphide. The equations for these two reactions are:
    H2S+O2SO2+H2OH2S+SO2S+H2OH2S+O2SO2+H2OH2S+SO2S+H2O
    (25)
    Balance these two equations.
    Click here for the solution

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