Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

40.5 grams of \(\mathrm{BaCO}_{3}\) are formed.

Step 1: Understanding the Chemical Reaction

When carbon dioxide \((\mathrm{CO}_{2})\) is passed through a solution of barium hydroxide \((\mathrm{Ba(OH)}_{2})\), barium carbonate \((\mathrm{BaCO}_{3})\) is formed. The balanced chemical equation for this reaction is: \[ \mathrm{Ba(OH)}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{BaCO}_{3} + \mathrm{H_{2}O} \] The equation tells us that 1 mole of \(\mathrm{Ba(OH)}_{2}\) reacts with 1 mole of \(\mathrm{CO}_{2}\) to produce 1 mole of \(\mathrm{BaCO}_{3}\).

Step 2: Calculating Moles of BaCO3 Formed

The reaction will produce moles of \(\mathrm{BaCO}_{3}\) equivalent to the moles of \(\mathrm{Ba(OH)}_{2}\) present, because \(\mathrm{CO}_{2}\) is in excess. We have 0.205 moles of \(\mathrm{Ba(OH)}_{2}\), so 0.205 moles of \(\mathrm{BaCO}_{3}\) will also be formed.

Step 3: Finding Molar Mass of BaCO3

The molar mass of \(\mathrm{BaCO}_{3}\) is calculated by summing up the atomic masses of barium (Ba), carbon (C), and three oxygens (O). The atomic masses are approximately: Ba = 137.33 g/mol, C = 12.01 g/mol, O = 16.00 g/mol. Thus, the molar mass of \(\mathrm{BaCO}_{3}\) = 137.33 + 12.01 + 3*16.00 = 197.34 g/mol.

Step 4: Calculating the Mass of BaCO3 Formed

Now, calculate the mass of \(\mathrm{BaCO}_{3}\) using the number of moles and its molar mass. Use the formula: \[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \] Substitute the known values: \[ \text{Mass of } \mathrm{BaCO}_{3} = 0.205 \times 197.34 = 40.4707 \text{ grams} \] Rounded to three significant figures, this is 40.5 grams.

Key Concepts

Stoichiometry

Stoichiometry is the foundational principle that allows chemists to quantitatively relate reactants and products in a chemical reaction. Its essence lies in the balanced chemical equation, which indicates the stoichiometric ratio - the proportion of molecules or moles that participate in or result from the reaction. In this exercise, the balanced reaction: \[ \mathrm{Ba(OH)}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{BaCO}_{3} + \mathrm{H_{2}O} \]helps us understand that one mole of barium hydroxide \((\mathrm{Ba(OH)}_{2})\) reacts with one mole of carbon dioxide \((\mathrm{CO}_{2})\) to produce one mole of barium carbonate \((\mathrm{BaCO}_{3})\) and water. Since this ratio is 1:1:1, stoichiometry allows us to deduce that if we start with 0.205 moles of \(\mathrm{Ba(OH)}_{2}\), then 0.205 moles of \(\mathrm{BaCO}_{3}\) will also be produced, assuming \(\mathrm{CO}_{2}\) is in excess. Understanding and applying stoichiometry is crucial in accurately predicting product amounts and determining limiting reagents in reactions.

Molar Mass Calculation

The concept of molar mass is crucial in converting between the mass of a substance and the amount of substance in moles, as defined by its chemical formula. Calculating molar mass involves summing the atomic masses of all atoms represented in the formula. For barium carbonate \((\mathrm{BaCO}_{3})\), we perform the following calculation:

• Barium (Ba) has an atomic mass of approximately 137.33 g/mol.
• Carbon (C) has an atomic mass of about 12.01 g/mol.
• Oxygen (O) has an atomic mass of around 16.00 g/mol, and since there are three oxygens in \(\mathrm{BaCO}_{3}\), we calculate 3 \(\times\) 16.00 = 48.00 g/mol for the oxygen atoms.

Adding these together: \[ 137.33 + 12.01 + 48.00 = 197.34 \, \text{g/mol} \]This resulting figure, 197.34 g/mol, is the molar mass of \(\mathrm{BaCO}_{3}\). Knowing this allows us to convert from moles of \(\mathrm{BaCO}_{3}\) to grams, and vice versa, enhancing our ability to handle real-world quantities in chemical reactions.

Barium Carbonate Formation

Barium carbonate formation in a laboratory setting involves a direct reaction between barium hydroxide \((\mathrm{Ba(OH)}_{2})\) and carbon dioxide \((\mathrm{CO}_{2})\). This reaction is a classic example of a precipitation reaction, where an insoluble solid, in this case, \(\mathrm{BaCO}_{3}\) forms and precipitates out of solution, balancing both mass and charge. When \(\mathrm{CO}_{2}\) is bubbled through a \(\mathrm{Ba(OH)}_{2}\) solution, the carbon dioxide interacts with barium ions \((\mathrm{Ba}^{2+})\) and hydroxide ions \((\mathrm{OH}^{-})\) in solution. The carbon dioxide reacts to form carbonate ions \((\mathrm{CO}_{3}^{2-})\), which then combine with the barium ions to form barium carbonate. This results in a white precipitate of \(\mathrm{BaCO}_{3}\) settling out of the solution.The formation of \(\mathrm{BaCO}_{3}\) is key in various industrial applications, such as in the production of ceramics and in the purification of water, illustrating both its chemical significance and practical importance.