Determine how many grams of N2 are produced from the reaction of 8.60 g of H2O2 and 5.76 g of N2H4

Question

2H2O2(l)+N2H4(l) ———> 4H₂O(g)+N₂(g)

Answer 1

2 H₂O₂ + N₂H₄ → 4 H₂O + N₂

First determine whether H₂O₂ or N₂H₄ is the limiting reagent by converting grams of each reactant to mol N₂. MW of H₂O₂ = 34.01 g/mol; MW of N₂H₄ = 32.05 g/mol.

8.60 g H₂O₂ x (1 mol H₂O₂ / 34.01 g H₂O₂) x (1 mol N₂ / 2 mol H₂O₂) = 0.126 mol N₂

5.76 g N₂H₄ x (1 mol N₂H₄ / 32.05 g N₂H₄) x (1 mol N₂ / 1 mol N₂H₄) = 0.180 mol N₂

From above, you can see that H₂O₂ produces less N₂, therefore, it is the limiting reagent. Convert 0.126 mol N₂ to g N₂ using the MW of N₂ = 28.00 g/mol.

0.126 mol N₂ x (28.00 g N₂ / 1 mol N₂) = 3.53 g N₂