Consider the reaction between N2H4 and N2O4: 2N2H4( g )+N2O4( g )3N2( g )+4H2O(g
ID: 960748 • Letter: C
Question
Consider the reaction between N2H4 and N2O4: 2N2H4(g)+N2O4(g)3N2(g)+4H2O(g)
A reaction vessel initially contains 23.5 gN2H4 and 74.9 g of N2O4.
Calculate the mass of N2H4 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Calculate the mass of N2O4 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Calculate the mass of N2 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Calculate the mass of H2O that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)
Explanation / Answer
Let's write again the reaction:
2N2H4(g)+N2O4(g)3N2(g)+4H2O(g)
Now, assuming 100% yield, we need to know who's the limiting reactant, let's calculate the moles of the reactants:
moles N2H4 = 23.5 / (2*14 + 4) = 0.7344 moles
moles N2O4 = 74.9 / (2*14 + 4*16) = 1.2483 moles
2 moles N2H4 ------> 1 mole N2O4
0.7344 moles -------> x
x = 0.7344 * 1 / 2 = 0.3672 moles N2O4
This means that in order to these compounds react, we need at least 0.3672 moles of N2O4, and we have 1.2483 moles, which mean that N2O4 is in excess and the N2H4 is the limiting reactant.
This also means that as N2H4 is the limiting reactant, this will be consumed at all, so, the mass left would be 0 g.
For N2O4, we know that is in excess and the remaining moles will be:
moles N2O4 after reaction = 1.2483 - 0.7344 = 0.5139 moles
mass N2O4 = 0.5139 * (60) = 30.834 g
The mass of N2, we know that the limiting reactant will produce the moles of the products so:
moles N2 = 0.7344 * 3/2 = 1.1016 moles
mass N2 = 1.1016 * 28 = 30.8448 g
The mass of water:
moles H2O = 0.7344 * 2 = 1.4688 moles
mass H2O = 1.4688 * 18 = 26.44 g
Hope this helps