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:
2 N2H4 (l) + N2O4 (l) à3N2 (g) + 4H2O (g)
If 290g of N2H4 and 116 g of N2O4 are mixed,
calculate the no. of moles of the excess reactant:
It appears you made a typo on the equation for the coefficient of N2 so I've corrected it. That funny a with an accent may have been an attempt at an arrow. Anyway, 2N2H4 (l) + N2O4 (l) ==> 3N2 (g) + 4H2O (g)
mols N2H4 = 290/32 = 9.06
mols N2O4 = 116/92 = 1.26
How many mols N2O4 are needed? That's
9.06 mols N2H4 x (1 mol N2O4/2 mols N2H4) = 4.53 and you have ONLY 1.26 mols N2O4 which isn't enough so N2O4 must be the limiting reagent (LR) while N2H4 is the excess reagent (ER). So how much N2H4 will the 1.26 mol N2O4 use? That's 1.26 mols N2O4 x (2 mols N2H4/1 mol N2O4) = 1.26*2 = 2.52.
mols ER = 9.06 initially - 2.52 used = ? mols N2H4 not used.