   Chapter 13, Problem 89QAP ### Introductory Chemistry: A Foundati...

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

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

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

# m>89. Ammonia and gaseous hydrogen chloride combine to form ammonium chloride.em>msp;  NH 3 ( g ) + HCl ( g )   NH 4 Cl ( s ) 4 . 2l L of NH 3 ( g ) at   27  ° C and 1 .0 2 mm is combined with 5 . 35 L of HCl ( g ) at   26  ° C and 0. 998 arm, what mass of NH 4 Cl ( s ) will be produced? Which gas is the limiting reactant? Which gas is present in excess?

Interpretation Introduction

Interpretation:

The mass of NH4Cl(s) produced should be calculated. The limiting reactant and gas present in excess should be determined. Concept Introduction:

According to ideal gas equation:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as 0.082 L atm K1 mol1.

Mass of gas can be calculated from number of moles and molar mass as follows:

m=n×M

Here, n is number of moles, m is mass and M is molar mass.

Explanation

Given Information:

The volume of NH3(g) is 4.21 L at temperature 27 C and pressure 1.02 atm. The volume of HCl(g) is 5.35 L at 26 C and pressure 0.998 atm.

Calculation:

The balanced chemical reaction is as follows:

NH3(g)+HCl(g)NH4Cl(s)

To calculate the limiting reactant, first calculate the number of moles of NH3 and HCl as follows:

n=PVRT

Convert the given temperature from C to K as follows:

0 C=273.15 K

Thus,

26 C=(26+273.15)K=299.15 K

And,

27 C=(27+273.15)K=300.15 K

Number of moles of NH3 will be:

n=(1.02 atm)(4.21 L)(0.082 L atm mol1 K1)(300

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