Problem 97 - Chemistry and Chemical Reactivity

Question

What mass of lime, CaO, can be obtained by heating \(125 \mathrm{kg}\) of limestone that is \(95.0 \%\) by mass \(\mathrm{CaCO}_{3} ?\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

66.51 kg of lime (CaO) can be obtained.

Step 1: Calculate the Mass of CaCO3 in the Limestone

Since the limestone is \(95.0\%\) \(\mathrm{CaCO}_{3}\), calculate the mass of \(\mathrm{CaCO}_{3}\) in the \(125 \text{ kg}\) of limestone. \[\text{Mass of } \mathrm{CaCO}_{3} = 125 \text{ kg} \times 0.95 = 118.75 \text{ kg}\]

Step 2: Determine the Molar Masses

Calculate the molar masses of \(\mathrm{CaCO}_{3}\) and \(\mathrm{CaO}\) using the periodic table:- Molar mass of \(\mathrm{CaCO}_{3}\): \(40.08 \text{ (Ca)} + 12.01 \text{ (C)} + 3 \times 16.00 \text{ (O)} = 100.09 \text{ g/mol}\).- Molar mass of \(\mathrm{CaO}\): \(40.08 \text{ (Ca)} + 16.00 \text{ (O)} = 56.08 \text{ g/mol}\).

Step 3: Convert Mass of CaCO3 to Moles

Convert the mass of \(\mathrm{CaCO}_{3}\) to moles using its molar mass:\[\text{Moles of } \mathrm{CaCO}_{3} = \frac{118,750 \text{ g}}{100.09 \text{ g/mol}} = 1186.44 \text{ mol}\]

Step 4: Calculate Moles of CaO Produced

From the balanced chemical equation, the mole ratio of \(\mathrm{CaCO}_{3}\) to \(\mathrm{CaO}\) is \(1:1\). Thus, the moles of \(\mathrm{CaO}\) produced are also \(1186.44 \text{ mol}\).

Step 5: Convert Moles of CaO to Mass

Convert moles of \(\mathrm{CaO}\) to mass using its molar mass:\[\text{Mass of } \mathrm{CaO} = 1186.44 \text{ mol} \times 56.08 \text{ g/mol} = 66,508.19 \text{ g}\]Convert grams to kilograms:\[66,508.19 \text{ g} = 66.51 \text{ kg}\]

Key Concepts

Chemical Reactions

In the world of chemistry, chemical reactions describe the process where reactants transform into products. This change involves breaking and forming chemical bonds. One kind of reaction is the decomposition reaction. It is what happens when we heat limestone, specifically calcium carbonate, \(\mathrm{CaCO}_3\). Here, calcium carbonate decomposes into calcium oxide, \(\mathrm{CaO}\), and carbon dioxide, \(\mathrm{CO}_2\).

The reaction can be represented as:

• \(\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})\)

• A solid to a solid and a gas
• Mass conserved during reaction

Here, we begin with one mole of \(\mathrm{CaCO}_3\) and end with one mole of \(\mathrm{CaO}\). This balanced equation tells us that there is a 1:1 molar ratio. It helps us predict the amount of products formed from given amounts of reactants.

Molar Mass

Molar mass, a crucial concept in chemistry, represents the mass of one mole of a substance. It's typically measured in grams per mole (g/mol) and is foundational in stoichiometry, which is the branch of chemistry dealing with the quantitative aspects of chemical reactions.

To find a substance's molar mass, sum the atomic masses of all atoms in its chemical formula. For example:

• Molar Mass of \(\mathrm{CaCO}_3\):
  • Calcium (Ca): \(40.08\, \mathrm{g/mol}\)
  • Carbon (C): \(12.01\, \mathrm{g/mol}\)
  • Oxygen (O): \(16.00\, \mathrm{g/mol}\) \(\times 3\)
  Total = \(100.09\, \mathrm{g/mol}\)

For \(\mathrm{CaO}\):

• • Calcium (Ca): \(40.08\, \mathrm{g/mol}\)
  • Oxygen (O): \(16.00\, \mathrm{g/mol}\)
  Total = \(56.08\, \mathrm{g/mol}\)

This information helps us convert between masses of material and moles, crucial for predicting the outcomes of chemical reactions.

Limestone Composition

Limestone primarily consists of the mineral calcium carbonate, \(\mathrm{CaCO}_3\). When an exercise like this asks us to consider the mass percent composition, it's simply detailing how much of the substance is calcium carbonate.

In this specific problem, the limestone is stated to be \(95.0\% \) \(\mathrm{CaCO}_3\). To understand how this composition figures into our calculations, we begin with the total mass of the limestone sample:

• Total mass given: 125 kg
• Calculate \(\mathrm{CaCO}_3\) mass:
  • \(125 \times 0.95 = 118.75 \text{ kg} \)

Recognizing this percentage allows us to understand how much pure \(\mathrm{CaCO}_3\) is present before the decomposition reaction proceeds.

Calcium Oxide Production

Producing calcium oxide, \(\mathrm{CaO}\), from limestone is an application of stoichiometry, specifically using the molar mass and mole concept.

The balanced equation shows conversion from \(\mathrm{CaCO}_3\) to \(\mathrm{CaO}\), and for every mole of \(\mathrm{CaCO}_3\), we produce a mole of \(\mathrm{CaO}\):

• Convert mass of \(\mathrm{CaCO}_3\) to moles:
  • \(\frac{118,750 \text{ g}}{100.09 \text{ g/mol}} = 1186.44 \text{ mol}\)
• Moles of \(\mathrm{CaO}\) produced: 1186.44 mol (1:1 Ratio)
• Convert to mass of \(\mathrm{CaO}\):
  • \(1186.44 \text{ mol} \times 56.08 \text{ g/mol} = 66,508.19 \text{ g}\)
  • Convert to kg: \(66.51 \text{ kg}\)

This result demonstrates the practical application of stoichiometry in calculating the yield of a product from a given reactant.