Question

What volume of \(\mathrm{CO}_{2}\), measured at STP is produced when \(27.5 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) is decomposed? $$ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$

Short answer

The volume of CO₂ produced when 27.5 g of CaCO₃ is decomposed at STP is 6.15 L.

Step-by-step solution

Calculate the moles of CaCO₃

First, we need to find the moles of CaCO₃. This can be done using the formula: $$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $$ The molar mass of CaCO₃ is \(40.08+12.01+3(16.00)=100.09 \mathrm{~g/mol}\). Given mass of CaCO₃ is \(27.5 \mathrm{~g}\), thus, $$ \text{moles of CaCO₃} = \frac{27.5 \mathrm{~g}}{100.09 \mathrm{~g/mol}} = 0.2748 \mathrm{~mol} $$

Use stoichiometry to calculate the moles of CO₂ produced

According to the balanced chemical equation, 1 mole of CaCO₃ will produce 1 mole of CO₂. Therefore, the moles of CO₂ produced can be calculated as: $$ \text{moles of CO₂ produced} = \text{moles of CaCO₃} = 0.2748 \mathrm{~mol} $$

Calculate the volume of CO₂ at STP

At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L. Using this information, we can find the volume of CO₂ produced: $$ \text{volume of CO₂} = \text{moles of CO₂} \times 22.4 \mathrm{~L/mol} = 0.2748 \mathrm{~mol} \times 22.4 \mathrm{~L/mol} = 6.15 \mathrm{~L} $$ So, the volume of CO₂ produced when 27.5 g of CaCO₃ is decomposed at STP is 6.15 L.

Key concepts

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. In the reaction you're looking at, calcium carbonate (\( \text{CaCO}_3 \)) decomposes to form calcium oxide and carbon dioxide. This is a simple decomposition reaction, where one compound breaks down into two or more products. Understanding the nature of chemical reactions helps you predict how various substances will interact and change.

• Reactants are the starting materials, like \( \text{CaCO}_3 \) in this reaction.
• Products are the substances formed, in this case, \( \text{CaO} \) and \( \text{CO}_2 \).
• Decomposition reactions break down a single compound into multiple simpler substances.

Chemical reactions are integral to both laboratory chemistry and natural processes, powering everything from industrial syntheses to biological metabolism.

Molar Mass

Molar mass is a crucial concept in chemistry that refers to the mass of one mole of a substance. It's expressed in grams per mole (g/mol) and is derived from the atomic masses of the elements in the compound. Knowing the molar mass allows us to convert between the mass of a substance and the number of moles, which is vital for understanding and performing chemical reactions.
In our example, the molar mass of calcium carbonate (\( \text{CaCO}_3 \)) is calculated as follows:

• Calcium (Ca) has an atomic mass of 40.08 g/mol.
• Carbon (C) has an atomic mass of 12.01 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol, and there are three oxygen atoms, making it 3 x 16.00 g/mol.

Therefore, the molar mass of \( \text{CaCO}_3 \) is 100.09 g/mol. By knowing this, we can accurately calculate the amount of substance and predict how much product will form in a reaction.

Gas Volume at STP

Gas volume at Standard Temperature and Pressure (STP) is a vital concept when dealing with gases in chemistry. STP is defined as a temperature of 0°C (273 K) and a pressure of 1 atm. Under these conditions, one mole of any ideal gas occupies a volume of 22.4 liters. This relationship, known as the molar volume of an ideal gas at STP, simplifies calculations involving gaseous products or reactants in chemical reactions.
In the decomposition of \( \text{CaCO}_3 \), the volume of \( \text{CO}_2 \) produced can be calculated using the molar volume:

• The moles of carbon dioxide produced: 0.2748 mol
• The volume of one mole of gas at STP: 22.4 L/mol

Simply multiply the moles of \( \text{CO}_2 \) by 22.4 L/mol to find the total volume, which in this case is 6.15 liters. This standard reference helps in comparing and predicting the behavior of gases in varying conditions.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants transform into products, using symbols and formulas to denote each substance involved. In a balanced chemical equation, the number of atoms for each element is the same on both sides of the equation, ensuring the conservation of mass in the reaction.
For the decomposition of calcium carbonate:\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]Here:

• The left side of the equation shows the reactant, \( \text{CaCO}_3 \).
• The right side of the equation shows the products, \( \text{CaO} \) and \( \text{CO}_2 \).
• The states of matter are indicated in parentheses, such as (s) for solid and (g) for gas.

Balancing chemical equations is essential for accurately understanding how much of each reactant is needed to produce a given quantity of product, and vice versa. It ensures that the proportions of substances are correct for the reaction to proceed efficiently.