/*! 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 113 You place \(2.56 \mathrm{g}\) of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 113

You place \(2.56 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) in a beaker containing \(250 .\) mL of \(0.125 \mathrm{M}\) HCl. When the reaction has ceased, does any calcium carbonate remain? What mass of \(\mathrm{CaCl}_{2}\) can be produced? $$\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$

Short Answer

Expert verified
Yes, 0.9975 g of CaCO3 remains. 1.733 g of CaCl2 can be produced.

Step by step solution

01

Calculate Moles of Reactants

First, we need to calculate the number of moles of \( \mathrm{CaCO}_{3} \) and \( \mathrm{HCl} \). The molar mass of \( \mathrm{CaCO}_{3} \) is approximately \( 100.09 \text{ g/mol} \). The moles of \( \mathrm{CaCO}_{3} \) are calculated as follows:\[\text{Moles of } \mathrm{CaCO}_{3} = \frac{2.56 \text{ g}}{100.09 \text{ g/mol}} = 0.0256 \text{ moles}\]For \( \mathrm{HCl} \), using the concentration and volume, we find its moles:\[\text{Moles of } \mathrm{HCl} = 0.125 \text{ M} \times 0.250 \text{ L} = 0.03125 \text{ moles}\]
02

Determine Limiting Reactant

Using the stoichiometry of the balanced equation \( \mathrm{CaCO}_{3} + 2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \), we see that 1 mole of \( \mathrm{CaCO}_{3} \) reacts with 2 moles of \( \mathrm{HCl} \). Thus, the complete reaction of 0.0256 mol of \( \mathrm{CaCO}_{3} \) would require:\[2 \times 0.0256 = 0.0512 \text{ moles of } \mathrm{HCl}\]Since we only have 0.03125 moles of \( \mathrm{HCl} \), \( \mathrm{HCl} \) is the limiting reactant, and \( \mathrm{CaCO}_{3} \) is in excess.
03

Determine Remaining \( \mathrm{CaCO}_{3} \)

Since \( \mathrm{HCl} \) is limiting, it will be consumed completely. Find the moles of \( \mathrm{CaCO}_{3} \) that reacted:\[\text{Moles of } \mathrm{CaCO}_{3} \text{ reacted} = \frac{0.03125}{2} = 0.015625 \text{ moles}\]The moles of \( \mathrm{CaCO}_{3} \) remaining are:\[0.0256 - 0.015625 = 0.009975 \text{ moles}\]Thus, some \( \mathrm{CaCO}_{3} \) remains.
04

Calculate Mass of \( \mathrm{CaCl}_{2} \) Produced

Calculate the moles of \( \mathrm{CaCl}_{2} \) produced, which is equal to the moles of \( \mathrm{CaCO}_{3} \) reacted, since they have a 1:1 ratio:\[0.015625 \text{ moles of } \mathrm{CaCl}_{2}\]Calculate the mass of \( \mathrm{CaCl}_{2} \) produced using its molar mass (110.98 g/mol):\[\text{Mass of } \mathrm{CaCl}_{2} = 0.015625 \times 110.98 = 1.733 \text{ g}\]

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 91影视!

Key Concepts

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

Limiting Reactant
When it comes to chemical reactions, identifying the limiting reactant is crucial. This is the reactant that runs out first, preventing the reaction from proceeding any further. In our exercise, the balanced chemical equation shows that one mole of \( \text{CaCO}_3 \) reacts with two moles of \( \text{HCl} \). Given 0.0256 moles of \( \text{CaCO}_3 \) and 0.03125 moles of \( \text{HCl} \), we need to determine which one limits the reaction.
  • 0.0256 moles of \( \text{CaCO}_3 \) requires 0.0512 moles of \( \text{HCl} \). However, only 0.03125 moles of \( \text{HCl} \) are available.
  • This means \( \text{HCl} \) is the limiting reactant, as there isn't enough to react with all the \( \text{CaCO}_3 \) present.

To find out if any \( \text{CaCO}_3 \) is left, it's important to calculate the amount that reacts. Since \( \text{HCl} \) is limiting, 0.015625 moles of \( \text{CaCO}_3 \) react, and 0.009975 moles are leftover. The presence of excess \( \text{CaCO}_3 \) confirms its non-limiting role in the reaction.
Chemical Reactions
Chemical reactions are processes where substances change to form new products. In our scenario, calcium carbonate (\( \text{CaCO}_3 \)) reacts with hydrochloric acid (\( \text{HCl} \)) to produce calcium chloride (\( \text{CaCl}_2 \)), carbon dioxide (\( \text{CO}_2 \)), and water (\( \text{H}_2\text{O} \)). Understanding the roles of each component is essential:
  • \( \text{CaCO}_3 \) acts as a base, reacting with the acid \( \text{HCl} \).
  • The products \( \text{CaCl}_2 \), \( \text{CO}_2 \), and \( \text{H}_2\text{O} \) signify a typical acid-carbonate reaction.

The balanced equation helps to ensure the conservation of mass, where the number of each type of atom on the reactant side equals the number on the product side. This stoichiometry involves paying attention to coefficients in the equation, dictating how many moles of each reactant are needed for the reaction. Understanding these principles is key to predicting the outcomes and leftovers in such reactions.
Molar Mass Calculation
Knowing how to calculate molar mass is a foundational step in stoichiometry. Molar mass is the mass of one mole of a substance and is expressed in grams per mole. It's crucial for converting between mass and moles, which is essential in determining limiting reactants and predicting product yields.
In this exercise, we have:
  • For \( \text{CaCO}_3 \), the molar mass is approximately 100.09 g/mol. It includes the atomic masses of calcium (Ca), carbon (C), and three oxygens (O).
  • For \( \text{CaCl}_2 \), the molar mass is approximately 110.98 g/mol.

By dividing the given mass by the molar mass, you can calculate the number of moles. For example, with 2.56 g of \( \text{CaCO}_3 \), by using the formula \( \frac{\text{mass}}{\text{molar mass}} \), we find it equals 0.0256 moles. This accuracy in calculating moles is critical in determining how much of a reactant will react or remain, as well as the amount of product formed.

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

Sulfuric acid can be prepared starting with the sulfide ore, cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right) .\) If each \(\mathrm{S}\) atom in \(\mathrm{Cu}_{2} \mathrm{S}\) leads to one molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what is the theoretical yield of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from \(3.00 \mathrm{kg}\) of \(\mathrm{Cu}_{2} \mathrm{S}\) ?

Antacids are chemical compounds that can give immediate relief from indigestion or heartburn because they contain carbonate or hydroxide ions that neutralize stomach acids. Some common active ingredients include \(\mathrm{NaHCO}_{3}, \mathrm{KHCO}_{3}\), \(\mathrm{CaCO}_{3}, \mathrm{Mg}(\mathrm{OH})_{2},\) and \(\mathrm{Al}(\mathrm{OH})_{3} .\) Although these compounds give quick relief, they are not recommended for prolonged consumption. Calcium carbonate may contribute to the growth of kidney stones, and calcium carbonate and aluminum hydroxide may cause constipation. Magnesium hydroxide, on the other hand, is a mild laxative that can cause diarrhea. Antacids containing magnesium, therefore, are often combined with aluminum hydroxide since the aluminum counteracts the laxative properties of the magnesium. (a) Which of the compounds listed above produce gas-forming reactions when combined with HCl? (b) One tablet of Tums Regular Strength Antacid contains \(500 . \mathrm{mg} \mathrm{CaCO}_{3}\) . (i) Write a balanced chemical equation for the reaction of \(\mathrm{CaCO}_{3}\) and stomach acid \((\mathrm{HCl})\) (ii) What volume (in mL) of 0.500 \(\mathrm{M}\) HCl(aq) will react completely with one tablet of Tums? (c) The active ingredients in Rolaids are \(\mathrm{CaCO}_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\). (i) Write a balanced chemical equation for the reaction of \(\mathrm{Mg}(\mathrm{OH})_{2}\) and \(\mathrm{HCl}\). (ii) If \(29.52 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl}\) is required to titrate one tablet of Rolaids and the tablet contains \(550 \mathrm{mg}\) of \(\mathrm{CaCO}_{3},\) what mass of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is present in one tablet? (d) Maalox may be purchased in either a liquid or solid form. One teaspoon of the liquid form of Maalox" contains a mixture of \(200 . \mathrm{mg}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(200 . \mathrm{mg}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} .\) What volume of 0.500 \(\mathrm{M} \mathrm{HCl}(\mathrm{aq})\) will react completely with one teaspoon of Maalox \(^{\pi / 2} ?\) (e) Which product neutralizes the greatest amount of acid when taken in the quantities presented above: one tablet of Tums" or Rolaids" or one teaspoon of Maalox"?

What is the hydronium ion concentration of a \(0.0013 \mathrm{M}\) solution of \(\mathrm{HNO}_{3} ?\) What is its \(\mathrm{pH} ?\)

Suppose you dilute \(25.0 \mathrm{mL}\) of a \(0.110 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to exactly \(100.0 \mathrm{mL}\). You then take exactly \(10.0 \mathrm{mL}\) of this diluted solution and add it to a 250-mL volumetric flask. After filling the volumetric flask to the mark with distilled water (indicating the volume of the new solution is \(250 . \mathrm{mL}\) ), what is the concentration of the diluted \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution?

The aluminum in a 0.764-g sample of an unknown material was precipitated as aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) which was then converted to \(\mathrm{Al}_{2} \mathrm{O}_{3}\) by heating strongly. If \(0.127 \mathrm{g}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is obtained from the 0.764-g sample, what is the mass percent of aluminum in the sample?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation
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.