Question

Consider the reaction between N2H4 and N2O4:
2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g)
A reaction vessel initially contains 27.5 gN2H4and 74.9 g of N2O4.

Calculate the mass of N2O4 that will be in the reaction vessel once the reactants have reacted as much as possible. (Assume 100% yield.)

Answer 1

2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g)

moles of N2H4 = mass / molar mass = 27.5 / 32 = 0.86 mol

moles of N2O4 = mass / molar mass = 74.9 / 92 = 0.81 mol

From the balanced equation,

1 mol of N2O4 reacts with 2 mol of N2H4.

moles of N2 can be formed from 0.86 mol of N2H4 = 0.86 x 3 / 2 = 1.29 mol

moles of N2 can be formed from 0.81 mol of N2O4 = 0.81 x 3 / 1 = 2.43 mol

Therefore N2H4 is limiting reactant.

moles of N2O4 required to form 1.29 mol of N2 = 1.29 x 0.81 / 2.43 = 0.43 mol

excess moles of N2O4 in the reaction vesssel = 0.81 - 0.43 = 0.38 mol

Mass of N2O4 in the reaction vesssel = 0.38 mol x 92 g/mol = 34.96 g <<<<--------(ANSWER)