The ability to convert a quantity of one substance into another within a chemical process forms the basis of chemical quantification. This calculation relies on stoichiometry, the method chemists use to measure relationships between reactants and products in a chemical reaction. Converting moles of nitrogen gas (\(\text{N}_2\)) into moles of ammonia (\(\text{NH}_3\)) requires a specific mathematical framework. The entire process hinges on using a balanced chemical equation to establish the exact proportion between the two substances.
The Essential Chemical Reaction
The production of ammonia from nitrogen and hydrogen gases is an industrial process known as the Haber process. Before any calculation can be performed, the chemical process must be represented by a balanced equation. The balanced chemical equation for this synthesis is \(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\). The numbers in front of each formula are stoichiometric coefficients, which ensure the number of atoms for each element is identical on both sides. This balancing step is the first requirement for any accurate stoichiometric calculation.
Deriving the Stoichiometric Mole Ratio
The coefficients from the balanced equation provide the precise molar relationship between every substance in the reaction. In the ammonia synthesis reaction, the coefficient of 1 for \(\text{N}_2\) and 2 for \(\text{NH}_3\) reveals the direct ratio between the two compounds. This means 1 mole of \(\text{N}_2\) will always produce 2 moles of \(\text{NH}_3\). This relationship is written as the conversion factor \(\frac{2\text{ mol } \text{NH}_3}{1\text{ mol } \text{N}_2}\), which allows for the cancellation of units during the calculation.
Step-by-Step Mole-to-Mole Conversion
The technique for performing this conversion is dimensional analysis, which involves multiplying the known quantity by the mole ratio. The starting point is the known quantity of nitrogen gas, expressed in moles. To ensure units cancel out properly, the unit being converted (\(\text{mol } \text{N}_2\)) must be placed in the denominator of the mole ratio. The desired unit (\(\text{mol } \text{NH}_3\)) remains in the numerator. The final result is obtained by multiplying the starting moles by the numerator (2) and dividing by the denominator (1).
Practical Example Calculation
Consider a scenario where a chemist starts with \(7.5\text{ mol } \text{N}_2\) and wants to determine the maximum amount of \(\text{NH}_3\) that can be produced. This quantity is multiplied by the specific mole ratio \(\frac{2\text{ mol } \text{NH}_3}{1\text{ mol } \text{N}_2}\) to set up the conversion. The arrangement is \(7.5\text{ mol } \text{N}_2 \times \frac{2\text{ mol } \text{NH}_3}{1\text{ mol } \text{N}_2}\). The units of \(\text{mol } \text{N}_2\) cancel out, leaving the desired unit of \(\text{mol } \text{NH}_3\). The final mathematical operation is \(7.5 \times 2\), meaning starting with \(7.5\text{ mol } \text{N}_2\) will produce \(15.0\text{ mol } \text{NH}_3\).