Question

Assuming fully decomposed, the volume of \(\mathrm{CO}_{2}\) released at STP on heating \(9.85 \mathrm{~g}\) of \(\mathrm{BaCO}_{3}\) (atomic mass, \(\mathrm{Ba}=137\) ) will be a. \(2.241\) b. \(4.961\) c. \(1.121\) d. \(0.841\)

Short answer

The volume of \( \text{CO}_2 \) released is 1.121 liters.

Step-by-step solution

Write the Chemical Equation

The decomposition reaction of barium carbonate \( \text{BaCO}_3 \) is as follows: \[ \text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2 \] One mole of \( \text{BaCO}_3 \) decomposes to release one mole of \( \text{CO}_2 \).

Calculate Molar Mass

Calculate the molar mass of \( \text{BaCO}_3 \): - Barium (Ba): 137 g/mol - Carbon (C): 12 g/mol - Oxygen (O): 16 g/mol for each oxygen, so 3 oxygen atoms give 48 g/mol Thus, the molar mass of \( \text{BaCO}_3 \) = 137 + 12 + 48 = 197 g/mol.

Determine Moles of \( \text{BaCO}_3 \)

Find the number of moles in 9.85 g of \( \text{BaCO}_3 \): \[ \text{Moles of } \text{BaCO}_3 = \frac{9.85}{197} \approx 0.05 \text{ moles} \]

Calculate Volume of \( \text{CO}_2 \) at STP

At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Thus, 0.05 moles of \( \text{CO}_2 \) will occupy: \[ \text{Volume of } \text{CO}_2 = 0.05 \times 22.4 = 1.12 \text{ L} \]

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It's based on the balanced equation of the reaction, providing a blueprint for the conservation of mass. In stoichiometry, the coefficients of the balanced equation tell us the relative amounts of moles of reactants and products. Think of it like a recipe; for every mole of one ingredient (or reactant), a certain number of moles of another ingredient (or product) is produced.

In the case of the decomposition of barium carbonate (\( \text{BaCO}_3 \)), the balanced chemical equation is: \[ \text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2 \] This equation tells us that one mole of \( \text{BaCO}_3 \) decomposes to produce one mole of \( \text{CO}_2 \). Here, the stoichiometry factor is 1:1, meaning they decompose and form at equal moles. This equality allows for straightforward calculations in determining the volume of gas produced from a given amount of reactant.
By using stoichiometry, we understand the direct relationship between the amount of reactant we start with and the amount of product formed in the reaction. It is a powerful bridge connecting the worlds of mass and volume through chemical equations.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It serves as a conversion factor between the mass of a substance and the moles of that substance. To find the molar mass of a compound, we simply add up the atomic masses of all the atoms in its formula.
Let's see this through the example of \( \text{BaCO}_3 \):

• Barium (Ba): Atomic mass = 137 g/mol.
• Carbon (C): Atomic mass = 12 g/mol.
• Oxygen (O): Atomic mass = 16 g/mol per atom, and in \( \text{BaCO}_3 \) there are three oxygen atoms, so 3 * 16 = 48 g/mol.

Summing these gives a molar mass for \( \text{BaCO}_3 \) of 137 + 12 + 48 = 197 g/mol.
This molar mass calculation is critical when converting from grams to moles, a step often needed to relate the mass of a substance to the amount in moles, which is used in chemical equations. Without this conversion factor, it would be challenging to assess how much of a substance reacts or is produced in a chemical reaction.

Gas Volume at STP

Standard temperature and pressure (STP) conditions are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm). Under these conditions, any ideal gas occupies a volume of 22.4 liters per mole. This relationship is a key component in calculating the volumes of gases produced or consumed in reactions at STP.
In the calculation, if we know how many moles of a gas are produced, we can directly determine its volume by multiplying the number of moles by 22.4 L/mol. This conversion makes it easy to understand how a small amount of gas in terms of moles translates into a larger, tangible volume in liters.
For instance, in the example of \( \text{BaCO}_3 \) decomposing, if we found that 0.05 moles of \( \text{CO}_2 \) are produced, we multiply by 22.4 L/mol to find the volume: \[ \text{Volume of } \text{CO}_2 = 0.05 \times 22.4 = 1.12 \text{ L} \] Thus, the use of STP conditions provides a simple and standardized way to connect the physical volumes of gases with chemical reactions.