   Chapter 10, Problem 87GQ

Chapter
Section
Textbook Problem

You are given 1.56 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl and O2,2 KClO3(s) → 2 KCl(s) + 3 O2(g)and 327 mL of O2 with a pressure of 735 mm Hg is collected at 19 °C. What is the weight percentage of KClO3 in the sample?

Interpretation Introduction

Interpretation: For the given 1.56g mixture of KClO3andKCl, the KClO3 gets decomposed to under given pressure conditions therefore, the weight percentage of KClO3 in the given sample should be determined.

Concept introduction:

Ideal gas Equation:

Any gas can be described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

nTPV = RnTPPV = nRTwhere,n = moles of gasP = pressureT = temperatureR = gas constant

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to the volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Van der Waal’s gas equation:

The van der Waal equation describes the ideal gas as it approaches to zero. The van der Waal equation contains correction terms a and b for the intermolecular forces and molecular size respectively.

The van der Waal equation is as follows,

[P+a(nV)2](Vnb)=RT

Explanation

Given,

Volume, V = 327 mL=0.327LTemperature, T=19oC=292.15Kmass = 1.56 gGas constant, R = 0.0821 L atm mol-1K-1Pressure, P = 735mm Hg =735mm Hg760mmHg=0.9671atm

Number of moles = PVRT=0

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