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Chapter 11: Problem 92

Upon heating, calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) decomposes to calcium oxide \((\mathrm{CaO})\) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) \begin{equation} \begin{array}{l}{\text { a. Determine the theoretical yield of } \mathrm{CO}_{2} \text { if } 235.0 \mathrm{g} \text { of }} \\\ {\mathrm{CaCO}_{3} \text { is heated. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. What is the percent yield of } \mathrm{CO}_{2} \text { if } 97.5 \mathrm{g} \text { of } \mathrm{CO}_{2} \text { is }} \\\ {\text { collected? }}\end{array} \end{equation}

Short Answer

Expert verified
The theoretical yield of CO2 is 103.34 g; the percent yield is 94.36%.

Step by step solution

01

Balanced Chemical Equation

Write the balanced chemical equation for the decomposition of calcium carbonate (\(\text{CaCO}_3\)) to calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)):\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]This equation indicates that 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\).
02

Molar Mass Calculation

Calculate the molar mass of \(\text{CaCO}_3\):- Calcium (Ca): 40.08 g/mol- Carbon (C): 12.01 g/mol- Oxygen (O): 16.00 g/mol\[ \text{Molar mass of } \text{CaCO}_3 = 40.08 + 12.01 + 3(16.00) = 100.09 \text{ g/mol} \]
03

Moles of CaCO3

Determine the number of moles of \(\text{CaCO}_3\) in 235.0 g:\[ \text{Moles of } \text{CaCO}_3 = \frac{235.0 \text{ g}}{100.09 \text{ g/mol}} = 2.348 \text{ mol} \]
04

Moles of CO2 Produced

Because the reaction produces 1 mole of \(\text{CO}_2\) per mole of \(\text{CaCO}_3\), 2.348 moles of \(\text{CaCO}_3\) will produce:\[ 2.348 \text{ mol of CO}_2 \text{ (theoretical)} \]
05

Mass of Theoretical CO2

Calculate the theoretical mass of \(\text{CO}_2\) using its molar mass:- Molar mass of \(\text{CO}_2\): (12.01 + 2 \times 16.00) g/mol = 44.01 g/mol\[ \text{Mass of } \text{CO}_2 = 2.348 \text{ mol} \times 44.01 \text{ g/mol} = 103.34 \text{ g} \]
06

Percent Yield Calculation

Given 97.5 g of \(\text{CO}_2\) is collected, the percent yield is calculated by:\[ \text{Percent Yield} = \left( \frac{97.5 \text{ g}}{103.34 \text{ g}} \right) \times 100\% = 94.36\% \]

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Key Concepts

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

Balanced Chemical Equation
In chemical reactions, a balanced chemical equation is crucial for understanding reactant-product relationships. It represents the conserved nature of mass and the rearrangement of atoms. For calcium carbonate (CaCO_3) decomposing to calcium oxide (CaO) and carbon dioxide (CO_2), the balanced equation is: \[\text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g)\]
  • This equation tells us that one molecule of CaCO_3 yields one molecule of CO_2.

This balance ensures the conservation of mass and atoms, which is foundational in chemical calculations. Thus, balancing chemical equations is the first essential step in solving stoichiometric problems.
Molar Mass
Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It serves as a bridge between the mass of a substance and the amount of substance (in moles). For our reaction, the molar masses are calculated as follows:
  • Calcium (Ca): 40.08 g/mol
  • Carbon (C): 12.01 g/mol
  • Oxygen (O): 16.00 g/mol

Using these values, you can find the molar mass of CaCO_3: \[ 40.08 + 12.01 + 3 \times 16.00 = 100.09 \text{ g/mol}\]
Understanding molar mass is vital as it allows you to convert between grams and moles, enabling further calculations in reaction stoichiometry.
Percent Yield
Percent yield quantifies the efficiency of a chemical reaction. It compares the amount of product actually obtained from a reaction to the theoretical maximum predicted by stoichiometry.
The formula to calculate percent yield is:
\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]
For the decomposition of CaCO_3, if 97.5 g of CO_2 is collected and 103.34 g is the theoretical amount generated, the percent yield is:
\[\left( \frac{97.5}{103.34} \right) \times 100\% = 94.36\%\]
A percent yield close to 100% indicates high efficiency, while lower values suggest potential losses or errors during the reaction.
Stoichiometry
Stoichiometry is the study of quantitative relationships between substances involved in chemical reactions. It allows chemists to predict how much reactant is needed and the amount of product formed.
In the decomposition reaction of CaCO_3 , stoichiometry helps us determine how much CO_2 will be generated from a given amount of CaCO_3 :
  • Firstly, calculate moles of CaCO_3 using its mass and molar mass.
  • Then, use the balanced equation to find the molar ratio, which tells us the moles of CO_2 produced.
This ensures precise calculations that are essential for practical applications in laboratories and industries.

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Most popular questions from this chapter

Film Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate \(\left(\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right)\) is produced. \begin{equation} \mathrm{AgBr}(\mathrm{s})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NaBr}(\mathrm{aq}) \end{equation} Determine the mass of \(\mathrm{Na}_{3} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\) produced if 0.275 \(\mathrm{g}\) of \(\mathrm{AgBr}\) is removed.

Antacid Fizz When an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate \(\left(\mathrm{NaHCO}_{3}\right),\) also called sodium bicarbonate, and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) \begin{equation} 3 \mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq}) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})\end{equation} How many moles of \(\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\) can be produced if one tablet containing 0.0119 \(\mathrm{mol}\) of \(\mathrm{NaHCO}_{3}\) is dissolved?

Can the percent yield of a chemical reaction be more than 100\(\% ?\) Explain your answer.

Apply Hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) decomposes to produce water and oxygen. Write a balanced chemical equation for this reaction, and determine the possible mole ratios.

Chrome The most important commercial ore of chromium is chromite \(\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right) .\) One of the steps in the process used to extract chromium from the ore is the reaction of chromite with coke (carbon) to produce ferrochrome (FeCr_ ). \begin{equation} 2 \mathrm{C}(\mathrm{s})+\mathrm{FeCr}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{FeCr}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \end{equation} What mole ratio would you use to convert from moles of chromite to moles of ferrochrome?
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