Chapter 9, Problem 90AP

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

Chapter
Section

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
175 views

Hydrazine N 2 H 4 , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine’s use as a fuel for rockets::math> N 2 H 4 ( l ) + O 2 ( g ) → N 2 ( g ) + 2 H 2 O ( g ) w many moles of each of the gaseous products are produced when 20.0 g of pure hydrazine ¡s ignited in the presence of 20.0 g of pure oxygen? How many grams of each product are produced?

Interpretation Introduction

Interpretation:

Number of moles of each gaseous product produces should be calculated and mass of each product should also be calculated.

Concept Introduction:

The expected or theoretical yield of the product depends on the limiting reactant of the reaction. Limiting reactant is the reactant that limits the mass of product formed in the reaction. The reactant other than limiting reactant is known as excess reactant.

Explanation

The balanced chemical reaction is as follows:

N2H4(l)+O2(g)N2(g)+2H2O(g)

To calculate the number of moles of gaseous product first calculate the limiting reactant as follows:

The number of moles of hydrazine can be calculated as follows:

n=mM

Molar mass of hydrazine is 32.0452 g/mol thus,

n=20 g32.0452 g/mol=0.6241 mol

From balanced chemical reaction, 1 mol of hydrazine gives 1 mol of nitrogen and 2 mol of water thus, number of moles of nitrogen and water will be 0.6241 mol and 2×0.6241 mol=1.248 mol respectively.

Calculate mass of nitrogen gas as follows:

m=n×M

Molar mass of nitrogen gas is 28 g/mol thus,

m=(0.6241 mol)(28 g/mol)=17.47 g

Similarly, molar mass of water is 18 g/mol thus, mass will be:

m=(1

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