Question

A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor

if 264g of N₂H₄ and 106g of N₂O₄ are mixed, calculate the number of moles of the excess reactant

Answer 1

2N₂H₄ + N₂O₄ → 3N₂ + 4H₂O

MM(N₂H₄) = 32.04 g/mol

n(N₂H₄) $= \frac{264}{32.04}=8.239 \;mol$

MM(N₂O₄) = 92.01 g/mol

n(N₂O₄) $= \frac{106}{92.01}=1.152 \;mol$

According to the reaction for each mole of N₂O₄ we need two moles of N₂H₄. So, N₂O₄ is a limiting reactant.

$n(N_2H_4)_{used} = 2n(N_2O_4) = 2 \times 1.152 = 2.304 \;mol$

The number of moles of the excess reactant (N₂H₄) = 8.239 -2.304 = 5.935 mol