/*! 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 35 \(5.00 \mathrm{g}\) of solid \(\... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 35

\(5.00 \mathrm{g}\) of solid \(\mathrm{CaCO}_{3}\) is thermally decomposed in the following reaction: \\[ \mathrm{CaCO}_{3}(\mathrm{s}) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \\] What mass of \(\mathrm{CaO}\) is formed? This oxide reacts with water to give \(\mathrm{Ca}(\mathrm{OH})_{2}\). Write a balanced equation for this process.

Short Answer

Expert verified
2.80 g of \(\text{CaO}\) is formed. The balanced reaction with water is \(\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2\).

Step by step solution

01

Write the Balanced Decomposition Reaction

The problem begins by asking us to decompose calcium carbonate (\(\text{CaCO}_3\)). The given reaction is: \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\) This equation is already balanced.
02

Calculate Molar Mass of the Compounds

To find the mass of \(\text{CaO}\) formed, we need to calculate the molar masses of \(\text{CaCO}_3\) and \(\text{CaO}\): - \(\text{CaCO}_3\): Calcium (Ca) is 40.08 g/mol, Carbon (C) is 12.01 g/mol, and 3 Oxygen (O) atoms each have a mass of 16.00 g/mol which add up to 48.00 g/mol. Total for \(\text{CaCO}_3\) is \(40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}\). - \(\text{CaO}\): Calcium is 40.08 g/mol and Oxygen is 16.00 g/mol. Total for \(\text{CaO}\) is \(40.08 + 16.00 = 56.08 \text{ g/mol}\).
03

Calculate Moles of \(\text{CaCO}_3\)

Using the given mass of \(\text{CaCO}_3\), calculate the moles: \[\text{moles of } \text{CaCO}_3 = \frac{5.00 \text{ g}}{100.09 \text{ g/mol}} \approx 0.04996 \text{ mol}\]
04

Use Stoichiometry to Find Moles of \(\text{CaO}\)

From the balanced equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CaO}\). Therefore, 0.04996 moles of \(\text{CaCO}_3\) will yield 0.04996 moles of \(\text{CaO}\).
05

Calculate the Mass of \(\text{CaO}\) Formed

Now, use the moles of \(\text{CaO}\) to find the mass: \[\text{mass of } \text{CaO} = 0.04996 \text{ mol} \times 56.08 \text{ g/mol} \approx 2.80 \text{ g}\] Thus, approximately 2.80 g of \(\text{CaO}\) is formed.
06

Write the Reaction of \(\text{CaO}\) with Water

The oxide \(\text{CaO}\) reacts with water to form calcium hydroxide (\(\text{Ca(OH)}_2\)). The balanced chemical equation for this reaction is: \(\text{CaO}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{Ca(OH)}_2(\text{s})\)

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Key Concepts

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

Stoichiometry
Stoichiometry is like the math toolbox in chemistry. It helps us relate the quantities of reactants and products in a chemical reaction. Imagine it as a chemical cookbook where the recipes (equations) tell you how much ingredient (molecules or moles) you need to make your desired dish (product).
  • Moles measure quantities in chemistry, similar to how dozens measure quantities in eggs. One mole contains Avogadro's number of particles.
  • When you look at a balanced equation, it shows the mole ratio of reactants to products, guiding you on how much of each substance you need.
In our example of decomposing calcium carbonate, stoichiometry lets us predict how much calcium oxide (CaO) results from a given amount of calcium carbonate (CaCO鈧�). By using the molar masses and the balanced equation, we can calculate these amounts precisely.
Thermal Decomposition
Thermal decomposition is an interesting chemical process where a compound breaks down into simpler substances when heated. Picture putting a block of ice into a warm skillet鈥攊t melts due to heat. Similarly, calcium carbonate (CaCO鈧�) breaks down when heated, releasing calcium oxide (CaO) and carbon dioxide (CO鈧�).
  • This process is useful in various industries. For example, making quicklime from limestone involves thermal decomposition.
  • When heated, a substance can lose or gain electrons, break bonds, and generally form more stable or energetically favorable compounds.
In the case of our exercise, the heat causes CaCO鈧� to separate into CaO and CO鈧�, a classic example of thermal decomposition.
Balanced Chemical Equations
Balanced chemical equations are like the grammar rules of chemistry, ensuring that what goes into a reaction is equal to what comes out. Think of them as the ultimate checklist for a reaction.
  • Each side of the equation must have the same number and type of atoms, satisfying the conservation of mass鈥攚hich says that matter is neither created nor destroyed.
  • Balancing an equation involves adding coefficients, numbers placed before a compound, to make the number of each type of atom equal on both sides.
In our given exercise, the decomposition equation: \[ \text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \]is balanced. The equation for the reaction of CaO with water is also balanced: \[ \text{CaO}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{Ca(OH)}_2(\text{s}) \] This ensures that all atoms are accounted for during the transformation.
Calcium Compounds
Calcium compounds are prevalent in both nature and industry. They're not just found in our bones, but are also key in building materials. Understanding their reactions is crucial in many chemical processes.
  • Calcium carbonate (CaCO鈧�) is a common form, used in chalk, limestone, and antacid tablets.
  • When heated, it decomposes into calcium oxide (CaO), also known as quicklime. CaO is a pivotal material in construction for making cement.
  • Adding water to CaO produces calcium hydroxide (Ca(OH)鈧�), known as slaked lime, which is used to make lime mortar and in water treatment plants.
This transformation from CaCO鈧� to CaO and finally to Ca(OH)鈧� showcases the versatility of calcium compounds in chemical reactions and everyday uses.

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Most popular questions from this chapter

What do you understand by each of the following terms: proton, electron, neutron, nucleus, atom, radical, ion, cation, anion, molecule, covalent bond, compound, isotope, allotrope?

Balance the following equation for the reaction of aqueous phosphoric acid with sodium hydroxide: \\[ \begin{array}{l} \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \\ \qquad \mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array} \\] \(15.0 \mathrm{cm}^{3}\) aqueous phosphoric acid of concentration \(0.200 \mathrm{mol} \mathrm{dm}^{-3}\) is added to \(50.0 \mathrm{cm}^{3}\) aqueous \(\mathrm{NaOH}\) of concentration \(2.00 \mathrm{mol} \mathrm{dm}^{-3} .\) Which reagent is in excess, and how many moles of this reagent remain unreacted?

Balance the following equations. (a) \(\mathrm{AgNO}_{3}+\mathrm{MgCl}_{2} \rightarrow \mathrm{AgCl}+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Pb}\left(\mathrm{O}_{2} \mathrm{CCH}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{PbS}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (c) \(\mathrm{BaCl}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4}+\mathrm{KCl}\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{KI} \rightarrow \mathrm{PbI}_{2}+\mathrm{KNO}_{3}\) (e) \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{CaCO}_{3}+\mathrm{H}_{2} \mathrm{O}\)

(a) Glucose, \(C_{x} H_{y} O_{z},\) contains \(40.00 \%\) C and \(6.71 \%\) H. What is the empirical formula of glucose? If the molecular mass of glucose is \(180.16,\) determine its molecular formula. (b) A fluoride of tungsten, \(\mathrm{WF}_{x},\) contains \(38.27 \%\) F. What is the oxidation state of tungsten in this compound?

The mass spectrum of molecular bromine shows three lines for the parent ion, \(\mathrm{Br}_{2}^{+} .\) The isotopes for bromine are \(_{35}^{79} \mathrm{Br}(50 \%)\) and \(_{35}^{81} \mathrm{Br}(50 \%) .\) Explain why there are three lines and predict their mass values and relative intensities. Predict what the mass spectrum of HBr would look like; isotopes of hydrogen are given in Section \(1.7 .\) (Ignore fragmentation; see Chapter \(10 .\)
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