/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 175 A sample of limestone was dissol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of limestone was dissolved in hydrochloric acid, and the carbon dioxide gas that evolved was collected. If a \(0.1662\) -g sample of limestone gave \(34.56 \mathrm{~mL}\) of dry carbon dioxide gas at \(745 \mathrm{mmHg}\) and \(21^{\circ} \mathrm{C}\), what was the mass percentage of \(\mathrm{CaCO}_{3}\) in the limestone?

Short Answer

Expert verified
The mass percentage of \(\text{CaCO}_3\) in the limestone is approximately 83.71%.

Step by step solution

01

Understand the Problem

The limestone sample is made of calcium carbonate, which reacts with hydrochloric acid to produce carbon dioxide. Our goal is to find the mass percentage of \( \text{CaCO}_3 \) in the sample after collecting the carbon dioxide produced.
02

Use the Ideal Gas Law

We need to calculate the moles of \(\text{CO}_2\) using the ideal gas law: \(PV=nRT\). First, convert the volume of the gas from \(\text{mL}\) to \(\text{L}\) (\(0.03456\, \text{L}\)), pressure from \(\text{mmHg}\) to \(\text{atm}\) using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\) (\(745/760 \approx 0.9803\, \text{atm}\)), and temperature from Celsius to Kelvin (\(21 + 273 = 294\, \text{K}\)).
03

Calculate Moles of CO2 Produced

Substitute the values into the ideal gas equation: \[ n = \frac{PV}{RT} = \frac{(0.9803)(0.03456)}{(0.0821)(294)} \approx 0.00139 \text{ moles of } \text{CO}_2 \]
04

Use Stoichiometry to Find Moles of CaCO3

In the reaction, \( \text{CaCO}_3 \rightarrow \text{CO}_2 + \text{other products}\), 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore, the moles of \(\text{CaCO}_3\) is also \(0.00139 \text{ moles}\).
05

Find Mass of CaCO3 in Limestone

Use the molar mass of \(\text{CaCO}_3\): \(100.0869 \text{ g/mol}\). Calculate the mass:\[ \text{Mass of } \text{CaCO}_3 = 0.00139 \times 100.0869 \approx 0.1391 \text{ g} \]
06

Calculate Mass Percentage of CaCO3

Finally, find the mass percentage of \(\text{CaCO}_3\) in the limestone sample using the formula:\[ \text{Mass percentage} = \left( \frac{\text{Mass of } \text{CaCO}_3}{\text{Total mass of sample}} \right) \times 100 = \left( \frac{0.1391}{0.1662} \right) \times 100 \approx 83.71\% \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a crucial tool in chemistry, helping us predict the behavior of gases. It is represented by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. The equation tells us how a gas behaves under different conditions.
When solving problems like the limestone sample exercise, you must convert units like pressure and temperature to those used in the Ideal Gas Law:
  • Pressure is often converted from mmHg to atm using the factor: \(1 \text{ atm} = 760 \text{ mmHg}\).
  • Temperature is converted from Celsius to Kelvin by adding 273.
Using the law helps us understand how many moles of gas are present in a sample, which leads to further calculations in experiments involving reactions.
Stoichiometry
Stoichiometry is the study of the relationships between amounts of reactants and products in chemical reactions. In our limestone problem, stoichiometry plays a role in determining how much \( ext{CaCO}_3\) is involved based on the \( ext{CO}_2\) evolved.
We look at mole ratios, which are based on the balanced chemical equation showing that \(1\) mole of \( ext{CaCO}_3\) yields \(1\) mole of \( ext{CO}_2\). This equality allows us to determine how much \( ext{CaCO}_3\) reacted using the moles of \( ext{CO}_2\) produced. The beauty of stoichiometry lies in converting measurable quantities to chemical equations, helping in predicting yields.
CaCO3
Calcium carbonate \(\text{CaCO}_3\) is a common substance found in rocks as the minerals calcite and aragonite, most notably in limestone. It is a compound made up of calcium, carbon, and oxygen.
This compound is essential in our exercise as it reacts with hydrochloric acid to release \( ext{CO}_2\), which we then measure to perform further calculations. Knowing the properties and reactions of \( ext{CaCO}_3\) helps in understanding how to quantify its presence in mixtures or solutions.
Limestone Sample
A limestone sample is primarily composed of calcium carbonate. In the exercise, a known mass of this sample was dissolved, and \( ext{CO}_2\) produced was collected.
The mass percentage calculation revolves around understanding how much \( ext{CaCO}_3\) is in that sample, given that not everything in the limestone will necessarily be \( ext{CaCO}_3\). The goal is to determine how pure the \( ext{CaCO}_3\) content in the sample is, which is crucial for various industrial and geological applications.
Mole Calculation
Mole calculations are vital for determining quantities in chemical reactions. The number of moles provides a link between the mass of a substance and its chemical quantity.
In our scenario, we used the Ideal Gas Law to find the moles of \( ext{CO}_2\). Knowing the moles of a gas helps predict and understand the extent of reactions because chemical equations are balanced in terms of moles. Leveraging mole calculations ensures precise understanding and capability in predicting reaction behaviors and quantities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When aluminum sulfate is dissolved in water, it produces an acidic solution. Suppose the \(\mathrm{pH}\) of this solution is raised by the dropwise addition of aqueous sodium hydroxide. (a) Describe what you would observe as the \(\mathrm{pH}\) continues to rise. (b) Write balanced equations for any reactions that occur.

Do you expect an aqueous solution of sodium hypochlorite to be acidic, neutral, or basic? What about an aqueous solution of sodium perchlorate?

Sulfur is formed in volcanic gases when sulfur dioxide reacts with hydrogen sulfide. The same reaction has been proposed as a method of removing sulfur dioxide from the gases emitted from coal-fired electricity generation plants. The emitted gases, containing sulfur dioxide, are bubbled through an aqueous solution of hydrogen sulfide. (a) Write the balanced chemical equation for the reaction of sulfur dioxide gas with hydrogen sulfide gas to produce solid sulfur and water vapor. (b) Suppose \(5.00 \mathrm{~L}\) of sulfur dioxide at \(748 \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\) reacts with \(15.1 \mathrm{~g}\) of hydrogen sulfide. How many grams of sulfur would be produced?

Sodium hydrogen sulfite is prepared from sodium carbonate and sulfur dioxide: \(\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(2 \mathrm{NaHSO}_{3}(a q)+\mathrm{CO}_{2}(g)\) How many grams of \(\mathrm{NaHSO}_{3}\) can be obtained from \(25.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Discuss the bonding in each of the following molecules or ions. What is the expected molecular geometry? a. \(\mathrm{HClO}\) b. \(\mathrm{ClO}_{4}^{-}\) c. \(\mathrm{ClF}_{5}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.