Balancing a chemical equation

The concept of balancing a chemical equation is based on the law of conservation of mass, where the total mass of reactants equals the total mass of products.

For example, the neutralisation reaction of sodium hydroxide with sulphuric acid could be:

$NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)$

However, you’ll realise that the number of atoms on the left hand side of the equation is not equal to that on the right, which means that the law of conservation of mass is violated. To balance the equation, we adjust the stoichiometric coefficients to give:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)\; \; \; \; \; \; \; \; 1$

$NaOH(aq)+\frac{1}{2}H_2SO_4(aq)\rightarrow \frac{1}{2}Na_2SO_4(aq)+H_2O(l)$

Since we usually represent the stoichiometric coefficients as whole numbers, eq1 is preferred.

Question

Construct and balance the chemical equation for the complete combustion of 1-heptene.

Answer

First, write a draft equation excluding the stoichiometric coefficients

$C_7H_{14}(l)+O_2(g)\rightarrow CO_2(g)+H_2O(l)$

Next, balance the equation by focusing on one element at a time. For example, we can start with carbon, followed by hydrogen and then oxygen to give the following equations:

$C_7H_{14}(l)+O_2(g)\rightarrow{\color{Red} \mathbf{7}}CO_2(g)+H_2O(l)$

$C_7H_{14}(l)+O_2(g)\rightarrow{\color{Red} \mathbf{7}}CO_2(g)+{\color{Red} \mathbf{7}}H_2O(l)$

$C_7H_{14}(l)+{\color{Red} \mathbf{\frac{21}{2}}}O_2(g)\rightarrow{\color{Red} \mathbf{7}}CO_2(g)+{\color{Red} \mathbf{7}}H_2O(l)\; \; \; \; \; \; \; \; 2$

Since whole numbers of stoichiometric coefficients are preferred, we multiply both sides of eq2 by 2 to give:

$2C_7H_{14}(l)+21O_2(g)\rightarrow 14CO_2(g)+14H_2O(l)$

Balancing a chemical equation

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