   Chapter 10, Problem 102IL

Chapter
Section
Textbook Problem

A mixture of NaHCO3 and Na2CO3 has a mass of 2.50 g. When treated with HCl(aq), 665 mL of CO2 gas is liberated with a pressure of 735 mm Hg at 25 °C. What is the weight percent of NaHCO3 and Na2CO3 in the mixture? (See Study Question 101 for the reactions that occur.)

Interpretation Introduction

Interpretation:

The percent composition by mass of Na2CO3 and NaHCO3 in the mixture should be calculated.

Concept Introduction:

Ideal gas Equation:

Any gas is 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 its 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.

Explanation

The balanced chemical equation for the reaction of the two given gases with HCl is as follows,

NaHCO3+HClNaCl+H2O+CO2Na2CO3+HCl2NaCl+H2O+CO2

First using the ideal gas equation the moles of CO2 produced is calculated as follows,

Here, we carry an extra significant figure throughout this calculation to limit rounding errors.

nCO2=PVRT =(0.967atm)(665×10-3L)(0.0821L.atmK.mol)(289.15)K=0.02709molCO2

From the balanced chemical equation it is clear that one mole of both the gases gives rise to one mole of CO2 hence, the total calculated amount of 0.02709molCO2 should be then produced by 0.02709mol of gases present in the given mixture.

Hence, there exists 1:1 mole ratio between CO2 and both reactants (Na2COand MgCO3), which show that 0.02709 of the mixture must have reacted.

molNa2CO3+molNaHCO3=0.02709mol

Let x be the mass of Na2CO3 in the mixture, then (2.50- x) is the mass of NaHCOin the mixture.    (xgNa2CO3×1molNa2CO3105.99gNa2CO3)+[(2.50-x)gMgCO3×1molNaHCO384.007gNaHCO3]=0.08436mol

9.4349×10-3x+(2.50-x)0.012 = 0.027099.435×10-3x+0.03-0.012x = 0.02709-2.565×10-3x = -3

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

A deficiency of niacin may result in which disease? pellagra beriberi scurvy rickets

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What is motion?

An Introduction to Physical Science

How do star clusters confirm that stars evolve?

Foundations of Astronomy (MindTap Course List)

Why is plastic so dangerous to marine organisms?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 