$$N_2 + 3H_2 \rightarrow 2NH_3$$
To determine the limiting reagent, we need to calculate the number of moles of each reactant and compare them to the stoichiometric ratio in the balanced equation.
1. Calculate the number of moles of nitrogen:
Moles of nitrogen = mass / molar mass
Moles of nitrogen = 150.0 g / 28.01 g/mol = 5.35 mol
2. Calculate the number of moles of hydrogen:
Moles of hydrogen = mass / molar mass
Moles of hydrogen = 32.1 g / 2.016 g/mol = 15.9 mol
3. Compare the number of moles of each reactant to the stoichiometric ratio:
According to the balanced equation, 1 mole of nitrogen reacts with 3 moles of hydrogen. Therefore, the stoichiometric ratio of nitrogen to hydrogen is 1:3.
- For nitrogen: 5.35 mol / 1 = 5.35
- For hydrogen: 15.9 mol / 3 = 5.3
4. Determine the limiting reagent:
The limiting reagent is the reactant that is present in a smaller mole ratio compared to the stoichiometric ratio. In this case, both nitrogen and hydrogen have the same mole ratio of 5.35. Therefore, there is no limiting reagent. This means that both nitrogen and hydrogen will be completely consumed in the reaction, and the reaction will go to completion.
