If the reaction N₂ + 3 H₂ → 2 NH₃ is carried out using 1.40 g of N₂ and 0.400 g of H₂,what mass of excess reactant will remain?

Question

Quizplus · Accepted Answer

We need to first determine which reactant is limiting and which is in excess. To do this, we can calculate the moles of each reactant: moles of N₂ = 1.40 g / (227.3 g/mol) = 0.00616 mol moles of H₂ = 0.400 g / (97.1 g/mol) = 0.00412 mol From the balanced chemical equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to form 2 moles of NH₃. This means that we only need 0.00205 mol of N₂ to fully react with the 0.00412 mol of H₂ we have. Since we have more than 0.00205 mol of N₂, it is not the limiting reactant. Therefore, the excess reactant is H₂. To determine the mass of excess reactant that remains, we can use the following calculation: moles of H₂ used = 0.00205 mol moles of H₂ initially = 0.00412 mol moles of H₂ remaining = 0.00412 mol - 0.00205 mol = 0.00207 mol Finally, we can convert the moles of remaining H₂ to grams: mass of H₂ remaining = 0.00207 mol x 97.1 g/mol = 0.201 g Therefore, the best answer is (C) 0.100 g H₂, which is half of the remaining mass of excess H₂.

If the Reaction N2 + 3 H2 → 2 NH3

If the Reaction N₂ + 3 H₂ → 2 NH₃