   Chapter 5, Problem 42QAP ### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373

#### Solutions

Chapter
Section ### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373
Textbook Problem

# A certain laser uses a gas mixture consisting of 9.00 g HCI, 2.00 g H2, and 165.0 g of Ne. What pressure is exerted by the mixture in a 75.0-L tank at 22°C? Which gas has the smallest partial pressure?

Interpretation Introduction

Interpretation:

The pressure exerted by the mixture in a 75.0 L tank at 22 C and the gas with the smallest partial pressure should be determined.

Concept introduction:

The Dalton’s Law of partial pressure of gases should be used to solve the problem. According to the law of partial pressure, the total pressure exerted in a mixture of non-reacting gases is equal to the sum of the partial pressures exerted by each individual gas.

Explanation

Before calculating the partial pressure first, we have to calculate the number of moles using the following formula:

n=massMM.. ..... (1)

In the above equation, let us substitute 9 g for mass and 36.5 g/mol for MM (molar mass) to find the number of moles of HCl

nHCl=9g36.5g/mol=0.25mol

In the above equation, let us substitute 165g for mass and 20g/mol for MM to find the number of moles of H2

nH2=2g2g/mol=1mol

In the above equation, let us substitute 165g for mass and 20g/mol for MM to find the number of moles of Ne

nNe=165g20g/mol=8.25mol

Now, the ideal law relation is calculated using the following formula:

PV=nRT

Here, P is the pressure of the system, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature. In the above equation let us substitute 295K for T, 0.25mol for n, 75L for V and 0.0821L atm/mol K for R to calculate the pressure of HCl.

PHCl×75L=0.25mol×0.0821Latm/mol.K×295KPHCl=0.25mol×0.0821Latm/mol.K×295K75L=0.081atm

In the above equation, let us substitute 295K for T, 1mol for n, 75L for V and 0.0821L.atm/mol.K for R to calculate the pressure of H2.

PH2×75L=1mol×0

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