Consider the reaction between N2H4 and N2O4:
2 N2H4(g) + N2O4(g) → 3 N2(g) + 4 H2O(g)
A reaction vessel initially contains 27.5 g N2H4 and 74.9 g of N2O4. Calculate the masses of N2H4, N2O4, N2, and H2O that will be in the reaction vessel after the reactants have reacted as much as possible. (Assume 100% yield.)
Here, we are going to calculate the masses of each reactant after the completion of the reaction.
2 N2H4(g) + N2O4(g) → 3 N2(g) + 4H2O(g)
Molar mass of N2H4(g) = 32.0452 g/mol
Molar mass of N2O4(g) =92.011 g/mol
Determination of the Limiting reagent.
0.858 moles of N2H4 will require 0.429 moles of N2O4
so all the N2H4 will be used up before the N2O4 runs out, the N2H4 is the limiting reactant.
Thus, we know least amount of reactant determines the limiting reagent.
Hence, the limiting reactant is N2H4.
Thus, complete N2H4 is used to form the product
Therefore, Mass of N2H4(g) = 0.0 g