0.7 Representing chemical change (Page 4/6)

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Balance the following equation:

{CH}_{4} + O_{2} \to {CO}_{2} + H_{2} O

Count the number of atoms of each element in the reactants and products
Reactants: $C = 1$ ; $H = 4$ and $O = 2$

Products: $C = 1$ ; $H = 2$ and $O = 3$
Balance the equation
If we add a coefficient of 2 to $H_{2} O$ , then the number of hydrogen atoms in the reactants will be 4, which is the same as for the reactants. The equation will be:
${CH}_{4} + O_{2} \to {CO}_{2} + 2 H_{2} O$
Check that the atoms balance
Reactants: $C = 1$ ; $H = 4$ and $O = 2$

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

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 $O_{2}$ , then we will increase the number of oxygen atoms in the reactants by 2. The new equation is:
${CH}_{4} + 2 O_{2} \to {CO}_{2} + 2 H_{2} O$

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.

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

Identify the reactants and the products and write their chemical formulae
The reactants are nitrogen ( $N_{2}$ ) and hydrogen ( $H_{2}$ ) and the product is ammonia ( $NH_{3}$ ).
Write the equation so that the reactants are on the left and products on the right of the arrow
The equation is as follows:
$N_{2} + H_{2} \to {NH}_{3}$
Count the atoms of each element in the reactants and products
Reactants: $N = 2$ and $H = 2$

Products: $N = 1$ and $H = 3$
Balance the equation
In order to balance the number of nitrogen atoms, we could rewrite the equation as:
$N_{2} + H_{2} \to 2 {NH}_{3}$
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 ( $H_{2}$ ), then the hydrogen atoms and the nitrogen atoms balance. The final equation is:
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$

In our bodies, sugar ( $C_{6} H_{12} O_{6}$ ) reacts with the oxygen we breathe in to produce carbon dioxide, water and energy. Write the balanced equation for this reaction.

Identify the reactants and products in the reaction and write their chemical formulae.
Reactants: sugar ( $C_{6} H_{12} O_{6}$ ) and oxygen ( $O_{2}$ )

Products: carbon dioxide ( ${CO}_{2}$ ) and water ( $H_{2} O$ )
Write the equation by putting the reactants on the left of the arrow and the products on the right
$C_{6} H_{12} O_{6} + O_{2} \to {CO}_{2} + H_{2} O$
Count the number of atoms of each element in the reactants and the number of atoms of each element in the products
Reactants: $C = 6$ ; $H = 12$ and $O = 8$

Products: $C = 1$ ; $H = 2$ and $O = 3$
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 ${CO}_{2}$ , the equation looks like this:
$C_{6} H_{12} O_{6} + O_{2} \to 6 {CO}_{2} + H_{2} O$

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

Products: $C = 6$ ; $H = 2$ and $O = 13$
Change the coefficients again to try to balance the equation.
Let's try to get the number of hydrogens the same this time.
$C_{6} H_{12} O_{6} + O_{2} \to 6 {CO}_{2} + 6 H_{2} O$

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

Products: $C = 6$ ; $H = 12$ and $O = 18$
Now we just need to balance the oxygen atoms.
$C_{6} H_{12} O_{6} + 12 O_{2} \to 6 {CO}_{2} + 6 H_{2} O$

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

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

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Source: OpenStax, Chemistry grade 10 [caps]. OpenStax CNX. Jun 13, 2011 Download for free at http://cnx.org/content/col11303/1.4

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