Question

unit 1, lc2 practice
balance the following chemical equation:

this is an example of a reaction.

Explanation:

Step1: Balance carbon atoms

Let the coefficient of \(C_3H_8\) be \(a = 1\). Since there are \(3\) carbon atoms in \(C_3H_8\), the coefficient of \(CO_2\) is \(b=3\) (because each \(CO_2\) has \(1\) carbon atom).

Step2: Balance hydrogen atoms

There are \(8\) hydrogen atoms in \(C_3H_8\). Each \(H_2O\) has \(2\) hydrogen atoms. So if the coefficient of \(C_3H_8\) is \(1\), the coefficient of \(H_2O\) is \(c = 4\) (since \(8\div2=4\)).

Step3: Balance oxygen atoms

The right - hand side has \(3\times2+4\times1=6 + 4=10\) oxygen atoms. Each \(O_2\) has \(2\) oxygen atoms. So the coefficient of \(O_2\) is \(d=\frac{10}{2}=5\).

For the reaction type, a combustion reaction is a reaction in which a substance reacts with oxygen to produce carbon dioxide and water (in the case of a hydrocarbon reacting with oxygen). Here \(C_3H_8\) (a hydrocarbon) reacts with \(O_2\) to produce \(CO_2\) and \(H_2O\), so it is a combustion reaction.

Answer:

\(1C_3H_8+5O_2
ightarrow3CO_2 + 4H_2O\), combustion reaction.