Question

The mass of \(\mathrm{BaCO}_{3}\) formed when excess of \(\mathrm{CO}_{2}\) is passed through a solution having \(0.205\) mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is: (a) \(40.5 \mathrm{~g}\) (b) \(20.25 \mathrm{~g}\) (c) \(81 \mathrm{~g}\) (d) \(4.05 \mathrm{~g}\)

Short answer

The mass of \( \text{BaCO}_3 \) formed is \( 40.5 \, \text{g} \). Answer: (a).

Step-by-step solution

Write the Chemical Equation

The first task is to write the balanced chemical equation for the reaction between barium hydroxide and carbon dioxide. The equation is: \[ \text{Ba(OH)}_2 + \text{CO}_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} \]This equation shows that one mole of \( \text{Ba(OH)}_2 \) reacts with one mole of \( \text{CO}_2 \) to form one mole of \( \text{BaCO}_3 \).

Determine Moles of BaCO3 Formed

From the balanced equation, it is clear that the reaction occurs in a 1:1 ratio. Therefore, 0.205 moles of \( \text{Ba(OH)}_2 \) will produce 0.205 moles of \( \text{BaCO}_3 \).

Calculate the Molar Mass of BaCO3

Barium carbonate \( (\text{BaCO}_3) \) consists of barium (Ba), carbon (C), and oxygen (O). Calculate the molar mass: - Ba: 137 g/mol - C: 12 g/mol - O: 16 g/mol, but there are three oxygens, so 48 g/mol \[ \text{Molar mass of BaCO}_3 = 137 + 12 + 48 = 197 \, \text{g/mol} \]

Calculate the Mass of BaCO3 Formed

Now, use the moles of \( \text{BaCO}_3 \) to find the mass. Multiply the moles by the molar mass:\[ \text{Mass of BaCO}_3 = 0.205 \, \text{moles} \times 197 \, \text{g/mol} = 40.385 \, \text{grams} \]Rounding this to the nearest significant figures gives \( 40.5 \, \text{grams} \).

Key concepts

Barium Carbonate Formation

Barium carbonate (\( \text{BaCO}_3 \)), a compound of barium, carbon, and oxygen, forms through a chemical reaction between barium hydroxide (\( \text{Ba(OH)}_2 \)) and carbon dioxide (\( \text{CO}_2 \)). This process is commonly seen in chemical laboratories where solutions of barium hydroxide are used. The reaction can be illustrated with the balanced chemical equation:
\[\text{Ba(OH)}_2 + \text{CO}_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O}\]This reaction is straightforward. It shows a one-to-one ratio between barium hydroxide and carbon dioxide. For every mole of barium hydroxide, one mole of barium carbonate is produced.
Understanding this ratio is crucial in predicting the amount of \( \text{BaCO}_3 \) formed in any given reaction. This process is significant in various chemical manufacturing and laboratory procedures.

Chemical Equations

Chemical equations represent reactions in the form of symbols and formulas, showing the reactants and products. They are essential for understanding how substances interact to form new compounds.
The equation in our context, \( \text{Ba(OH)}_2 + \text{CO}_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} \), clearly demonstrates how barium hydroxide and carbon dioxide react to form barium carbonate and water. Each side of the equation must contain the same number of each type of atom, maintaining the law of conservation of mass.
Let's break down what this actually represents:

• Reagents are balanced quantitatively, meaning their coefficients (or numbers in front of them) are as low as possible.
• The phases of each compound can also be indicated (solid, liquid, gas, aqueous), adding further contextual understanding.
• Chemical equations are both theoretical and practical guidelines in predicting the outcome of a reaction, ensuring that we use the appropriate amount of reactants for the desired yield.

Knowing how to write and balance chemical equations is a foundational skill in chemistry, aiding both academic study and practical applications.

Molar Mass Calculation

Molar mass is a critical concept that aids in converting between moles and grams of a substance. It involves summing the atomic masses of each element in a compound, as represented by the formula.
For barium carbonate (\( \text{BaCO}_3 \)), the molar mass is calculated as follows:

• Barium (Ba) contributes 137 g/mol, being the heaviest component.
• Carbon (C) adds 12 g/mol.
• Oxygen (O), with three atoms present in \( \text{BaCO}_3 \), contributes 48 g/mol (since 16 g/mol per oxygen atom).

Adding these values yields the total molar mass:
\[\text{Molar mass of BaCO}_3 = 137 + 12 + 48 = 197 \text{ g/mol}\]With this, you can perform conversions necessary to determine how much product forms in a reaction. For example, if you know the number of moles of \( \text{BaCO}_3 \), multiplying by the molar mass gives you the mass in grams.
This calculation is used in various quantitative analyses and chemical computations. Having a solid understanding of how to calculate molar mass ensures accurate, reliable results in both theoretical and experimental chemistry.