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

Option c, 1.121 L of \( \mathrm{CO}_{2} \) is released.

Step-by-step solution

Understanding the Problem

We need to find the volume of carbon dioxide, \( \mathrm{CO}_{2} \), produced when \( 9.85 \mathrm{~g} \) of barium carbonate, \( \mathrm{BaCO}_{3} \), is heated and fully decomposed, measured in liters at standard temperature and pressure (STP). The atomic weight of barium (\( \mathrm{Ba} \)) is given as \( 137 \).

Writing the Decomposition Equation

The decomposition reaction of \( \mathrm{BaCO}_{3} \) is: \[ \mathrm{BaCO}_{3(s)} \rightarrow \mathrm{BaO}_{(s)} + \mathrm{CO}_{2(g)} \] This tells us that 1 mole of \( \mathrm{BaCO}_{3} \) produces 1 mole of \( \mathrm{CO}_{2} \).

Calculating Molar Mass of \( \mathrm{BaCO}_{3} \)

The molar mass of \( \mathrm{BaCO}_{3} \) is calculated by adding the atomic masses of barium (\( 137 \)), carbon (\( 12 \)), and three oxygens (\( 3 \times 16 \)): \[ 137 + 12 + 48 = 197 \text{ g/mol} \]

Finding Moles of \( \mathrm{BaCO}_{3} \)

We calculate the number of moles of \( \mathrm{BaCO}_{3} \) using its mass and molar mass: \[ \text{Moles of } \mathrm{BaCO}_{3} = \frac{9.85 \text{ g}}{197 \text{ g/mol}} \approx 0.05 \text{ moles} \]

Determining Moles of \( \mathrm{CO}_{2} \) Produced

Since the reaction is \( 1:1 \), \( 0.05 \) moles of \( \mathrm{BaCO}_{3} \) decompose to produce \( 0.05 \) moles of \( \mathrm{CO}_{2} \).

Calculating Volume of \( \mathrm{CO}_{2} \) at STP

At standard temperature and pressure (STP), 1 mole of gas occupies \( 22.4 \text{ L} \). Thus, the volume of \( \mathrm{CO}_{2} \): \[ \text{Volume of } \mathrm{CO}_{2} = 0.05 \text{ moles} \times 22.4 \text{ L/mol} = 1.12 \text{ L} \]

Final Answer

The volume of \( \mathrm{CO}_{2} \) released at STP is approximately \( 1.121 \text{ L} \).

Key concepts

Stoichiometry

Stoichiometry is the part of chemistry that helps us figure out the quantities of reactants and products in chemical reactions. It relies on a balanced chemical equation to determine the relationship between the amount of substances involved in the reaction.
In our problem about barium carbonate (\( \mathrm{BaCO}_3 \)), stoichiometry tells us that for every mole of \( \mathrm{BaCO}_3 \) that decomposes, one mole of carbon dioxide (\( \mathrm{CO}_2 \)) is produced. This is because the chemical equation is balanced and shows a 1:1 ratio between \( \mathrm{BaCO}_3 \) and \( \mathrm{CO}_2 \). Knowing this allows us to calculate how much \( \mathrm{CO}_2 \) we get from a known quantity of \( \mathrm{BaCO}_3 \).

• This step in stoichiometry involves balancing the chemical equation.
• Using molar ratios from the balanced equation helps us predict the amounts of products.

Understanding stoichiometry is crucial as it ensures that we know exactly how much product to expect from a given reactant under specified conditions.

Molar Mass Calculation

Molar mass is the mass of a given substance (element or compound) divided by the amount of substance. It is usually expressed in grams per mole (g/mol). Calculating the molar mass is an important step in solving many stoichiometry-related problems.
For our compound, barium carbonate (\( \mathrm{BaCO}_3 \)), we need to calculate its molar mass to figure out how many moles are present in a given sample. The molar mass is the sum of the atomic masses of all the atoms in the formula:

• Barium (\( \mathrm{Ba} \)): 137 g/mol
• Carbon (\( \mathrm{C} \)): 12 g/mol
• Oxygen (\( \mathrm{O} \)): 3 \( \times \) 16 g/mol = 48 g/mol

So, the total molar mass of \( \mathrm{BaCO}_3 \) is \( 137 + 12 + 48 = 197 \text{ g/mol} \).
With this information, we can calculate the moles of \( \mathrm{BaCO}_3 \) by dividing the mass of the sample by the molar mass. Having the moles helps in determining the amount of \( \mathrm{CO}_2 \) produced.

Ideal Gas Law

The Ideal Gas Law is an equation of state for a gas.This law combines several different gas laws and helps us understand how a gas behaves under different conditions of temperature, pressure, and volume. However, in standard temperature and pressure (STP) conditions, we know from empirical data that one mole of any gas occupies a volume of 22.4 liters.
This simplifies calculations like those in our exercise since we don't need to individually apply the Ideal Gas Law equations. Instead, we can take advantage of this approximation. In our problem, knowing the amount of \( \mathrm{CO}_2 \) moles produced allows us to directly calculate its volume at STP by multiplying the number of moles by 22.4 L/mol.

• This approach saves time and reduces complex computations.
• Remember that this is specifically under STP, which is 0 degrees Celsius and 1 atmosphere of pressure.

So, in this example, \( 0.05 \) moles of \( \mathrm{CO}_2 \) would equal \( 1.12 \) liters, as calculated from \( 0.05 \times 22.4 \text{ L} \). This helps us understand the relationship between the number of gas moles and their volume in real-life conditions.