Answer:
Step-by-step explanation:
To identify the limiting and excess reagents, we need to compare the moles of each reactant and determine which one is present in a stoichiometrically lesser amount.
Given:
Mass of silicon (Si) = 40.0 g
Mass of nitrogen (N2) = 25.5 g
To determine the moles of each reactant, we need to divide their masses by their respective molar masses.
Molar mass of silicon (Si) = 28.0855 g/mol
Molar mass of nitrogen (N2) = 28.0134 g/mol
Moles of silicon (Si) = 40.0 g / 28.0855 g/mol ≈ 1.424 mol
Moles of nitrogen (N2) = 25.5 g / 28.0134 g/mol ≈ 0.911 mol
Next, we need to determine the stoichiometric ratio of the reactants based on the balanced chemical equation. Let's assume the balanced equation is:
3Si + 2N2 → Si3N4
From the balanced equation, we can see that 3 moles of silicon react with 2 moles of nitrogen to form 1 mole of silicon nitride.
Comparing the mole ratio of silicon to nitrogen, we have:
1.424 mol Si : 0.911 mol N2
Since the ratio is not a whole number ratio, we can simplify it by dividing both sides by 0.911:
1.424 mol Si / 0.911 mol N2 ≈ 1.56
Based on this ratio, we can see that the stoichiometric ratio of silicon to nitrogen is approximately 1.56:1. This means that 1.56 moles of silicon react with 1 mole of nitrogen.
Now, comparing the given moles of silicon and nitrogen to the stoichiometric ratio, we can determine the limiting and excess reagents.
Since we have 1.424 mol of silicon and the stoichiometric ratio requires only 1.56 moles of silicon per mole of nitrogen, silicon is the limiting reagent.
Nitrogen, with 0.911 mol, is in excess because it is not fully consumed in the reaction.
In conclusion:
Silicon (Si) is the limiting reagent.
Nitrogen (N2) is the excess reagent.