If 2.29 kg of hydrazine are used in a rocket engine and 3.14 kg of oxygen are available for reaction,

Determine the limiting reactant, amount of reactant in excess, mass of each product formed

Reaction of hydrazine is

N2H4 + O2 - -> N2 + 2H2O

Mass of hydrazine mole = 14x2 + 1x4 = 32g/Mol

Mass of O2 mole = 16x2 = 32g/Mol

Number of moles in 2.29kg of hydrazine

2290/32=71.56Moles

1mol of hydrazine required to react with 1mol of O2

Amount of O2 required to react with hydrazine

71.56*32=2290

Required O2 amount is exactly the same.

Therefore limiting reactant is hydrazine and excess of 850g of O2 will be in excess.