Question

Some rocket engines use a mixture of hydrazine,N2H4 and hydrogen peroxide ,H2O2 as the propellent . the reaction is given by the following equation N2H4 + 2H2O gives N2 GAS + 4 H2O .how much of the excess reactant,remains unchanged? when 0.850 mol of N2H4 is mixed with 17 g of H2O2 ?

Answer 1

Hi, please consider the answer given below;

So what you can do in this case is to determine the number of moles of hydrogen peroxide and this will enable you to determine the reagent which was in excess.

So the number of moles of hydrogen peroxide= mass/molar mass of hydrogen peroxide

moles= 17/34

moles = 0.5 moles

Therefore from this it is evident that hydrazine is in excess

excess moles= 0.85 - 0.5 = 0.35 moles

mass= moles x molar mass of hydrazine

mass = 0.35 x 32

mass = 11.2 grams

Answer 2

Given:

No of moles of N2H4 = 0.85

Mass of H2O2 = 17 gm

To Find:

The mass of excess reactants remained unchanged after the reaction.

Calculation:

- The no of moles of H2O2 = 17/34 = 0.5 moles

- According to the given reaction:

N2H4 required to react with 2 moles of H2O2 = 1 mole

⇒ N2H4 required to react with 0.5 moles of H2O2 = (1/2) × 0.5 = 0.25 moles

- So, N2H4 is the limiting reagent here.

- The no of moles of N2H4 left unreacted = 0.85 - 0.25 = 0.60 moles

- The mass of N2H4 left unreacted = 0.6 × 32 = 19.2 gm

- So, 0.6 moles or 19.2 gm of excess N2H4 remains unchanged in the given conditions.