If 25.0 g N2, and 18.75 g H2, are used for this reaction, How much of the excess reactant will be left over?
To answer this question, we need to determine which reactant is limiting and which is in excess.
1. Write the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
2. Calculate the moles of each reactant:
Moles of N2 = 25.0 g / 28.02 g/mol = 0.892 moles
Moles of H2 = 18.75 g / 2.016 g/mol = 9.30 moles
3. Determine the limiting reactant by looking at the mole ratio in the balanced chemical equation. In this case, it is N2 because it has fewer moles compared to the H2 in the ratio (1:3).
4. Calculate the amount of NH3 produced from the limiting reactant:
0.892 moles N2 x (2 moles NH3 / 1 mole N2) = 1.784 moles NH3
5. Calculate the amount of NH3 that would be produced from the H2 if it were the limiting reactant:
9.30 moles H2 x (2 moles NH3 / 3 moles H2) = 6.20 moles NH3
Since the N2 is limiting the reaction, all of the H2 will be used up. Therefore, there will be no excess H2 left over.