#Rules
A balanced chemical equation is an algebraic representation that shows the relative number of molecules (or moles) of each substance involved in a reaction. Ensure the same number of atoms for each element on both sides of the equation, as required by the law of conservation of mass.
$$ \text{NH}_3 + \text{O}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} $$#To solve it
- Assign variables to represent the coefficients of each substance.
- Write one equation for each element, ensuring the total number of atoms is the same on both sides.
- Solve the equations to find the values of the variables.
- Adjust to the smallest whole-number integer combination of variables to represent the reaction.
#Example 1
Find balanced chemical equation for this reaction
$$ \text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O} $$#Solution:
$$ x\text{H}_2 + y\text{O}_2 \rightarrow z\text{H}_2\text{O} \quad x, y, z \text{ unknowns} \in \mathbb{Z} \quad \min(\text{combination}(x, y, z)) $$ Hydrogen: (2x = 2z)Oxygen: (2y = z) $$ \begin{cases} 2x - 2z = 0 \\ 2y - z = 0 \end{cases} $$ $$ \begin{pmatrix} 2 & 0 & -2 \\ 0 & 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Using RREF: $$ \begin{pmatrix} 2 & 0 & -2 & | & 0 \\ 0 & 2 & -1 & | & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & -1 & | & 0 \\ 0 & 1 & -\frac{1}{2} & | & 0 \end{pmatrix} $$ $$ x - z = 0 \quad \Rightarrow \quad x = z $$ $$ y - \frac{1}{2}z = 0 \quad \Rightarrow \quad y = \frac{1}{2}z $$ Solutions: $((x, y, z) = (z, \frac{1}{2}z, z) = (c, \frac{1}{2}c, c) \rightarrow \min(x, y, z) = (2, 1, 2))$ $$ \boxed{2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}} $$
#Example 2
Combustion of ammonia in oxygen produces nitrogen and water. Find balanced chemical equation for this reaction.
$$ \text{NH}_3 + \text{O}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} $$ $$ w\text{NH}_3 + x\text{O}_2 \rightarrow y\text{N}_2 + z\text{H}_2\text{O} $$Nitrogen: $(w = 2y)$ Oxygen: $(2x = z)$ Hydrogen: $(3w = 2z)$
Equations:
$$ \begin{cases} w - 2y = 0 \\ 2x - z = 0 \\ 3w - 2z = 0 \end{cases} $$ Matrix form: $$ \begin{pmatrix} 1 & 0 & -2 & 0 & | & 0 \\ 0 & 2 & 0 & -1 & | & 0 \\ 0 & 0 & 6 & -2 & | & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & -\frac{2}{3} & | & 0 \\ 0 & 2 & 0 & -1 & | & 0 \\ 0 & 0 & 6 & -2 & | & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & -\frac{2}{3} & | & 0 \\ 0 & 1 & 0 & -\frac{1}{2} & | & 0 \\ 0 & 0 & 1 & -\frac{1}{3} & | & 0 \end{pmatrix} $$ Solutions: $$ (w, x, y, z) = \left(\frac{2}{3}z, \frac{1}{2}z, \frac{1}{3}z, z\right) = \left(\frac{2}{3}c, \frac{1}{2}c, \frac{1}{3}c, c\right), \quad c \in \mathbb{R} $$ Minimum integer combination: $$ (w, x, y, z) = (4, 3, 2, 6) $$ Balanced equation: $$ \boxed{4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O}} $$
Areas and Volumes