190k views
2 votes
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

User Flyingace
by
7.9k points

1 Answer

4 votes

Answer:

Step-by-step explanation:

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.

User Clearlight
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.