If 25.0g N2, and 18.75g H2, are used for this reaction, which reactant will be the limiting reactant?
To determine the limiting reactant, we need to calculate the moles of each reactant and then use the balanced chemical equation to see which reactant will be used up first.
1. Calculate the moles of N2:
Molar mass of N2: 14.01 g/mol
Moles of N2 = 25.0g / 14.01 g/mol = 1.785 mol
2. Calculate the moles of H2:
Molar mass of H2: 2.016 g/mol
Moles of H2 = 18.75g / 2.016 g/mol = 9.292 mol
3. Now we need to determine which reactant will be the limiting reactant. To do this, we look at the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
From the balanced equation, it can be seen that one mole of N2 reacts with 3 moles of H2. Therefore, the molar ratio of N2 to H2 is 1:3.
Using this ratio, we see that 1.785 mol of N2 would require 1.785 mol x 3 H2 = 5.355 mol of H2. Since we only have 9.292 mol of H2 available, there is excess H2 present.
Therefore, the limiting reactant in this reaction is N2.