   # A 20.0−L stainless steel container at 25°C was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at 25°C? If the exact same experiment were performed, but the temperature was 125°C instead of 25°C, what would be the pressure in the tank? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 134AE
Textbook Problem
605 views

## A 20.0−L stainless steel container at 25°C was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at 25°C? If the exact same experiment were performed, but the temperature was 125°C instead of 25°C, what would be the pressure in the tank?

Interpretation Introduction

Interpretation: The pressure of given reaction tank at125oCshould be identified.

Concept introduction:

• Number of moles of a substance

According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

• Balance equation of a reaction is written according to law of conservation of mass.
• Mole ratio between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
• Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.
• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.
• Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.
• According to ideal gas law: the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

Pn=RTV , P and n are proportional.

Mathematically, for two gas mixtures (1 and 2) at constant V and T,

P1n1=P2n2P1P2=n1n2

### Explanation of Solution

To find: the number of moles ofH2and O2 in the given reaction.

The number of moles of H2is2VRTmol.

The number of moles of O2is3VRTmol.

The partial pressure of H2in the given reaction is2atm.

The partial pressure of O2in the given reaction is3atm.

The equation for finding number of moles of a substance, According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

The volume, temperature and R-value are constant for both H2andO2in the given reaction.

Hence,

Therefore,

The number of moles of H2in the given reaction is,

nH2=2VRTmol

The number of moles of O2in the given reaction is,

nO2=3VRTmol

Balanced chemical equation for the given reaction.

2H2+O22H2O

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

To find: the limiting reagent in the given reaction.

The limiting reagent in the given reaction is H2.

The number of moles of H2is calculated as2VRTmol.

The number of moles of O2is calculated as3VRTmol.

The balanced equation of given reaction is calculated as

2H2+O22H2O

Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

Assuming H2is limiting:

In balanced equation of given reaction, 2 mole of H2 is reacting with 1 mole ofO2.

That means,

2VRTmolH2×1molO22molH2=1VRTmolO2

Assuming isO2 liming:

In balanced equation of given reaction, 2 mole of H2 is reacting with 1 mole ofO2.

That means,

3VRTmolO2×2molH21molO2=6VRTmolH2

Because H2is having2VRTmol only in the given reaction, so H2 is the limiting reagent in the given reaction

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