Question

Volume of \(CO_2\) obtained by the complete decomposition of \(9.85\ gm \ BaCO_3\) is:

• A. \(2.24 \ lit.\)
• B. \(1.12\ lit.\)
• C. \(0.84\ lit.\)
• D. \(0.56\ lit.\)

Answer 1

The Correct Option is B

To solve this problem, we can follow these steps:

1. First, we need to understand the chemical reaction involved. The decomposition of BaCO_3 (Barium Carbonate) can be represented as follows:

BaCO_3 \rightarrow BaO + CO_2

2. This reaction indicates that 1 mole of Barium Carbonate decomposes to give 1 mole of Carbon Dioxide.
3. Next, we calculate the molar mass of BaCO_3:

Ba: 137 \ g/mol, \ C: 12 \ g/mol, \ O: 16 \times 3 = 48 \ g/mol

\text{Molar mass of } BaCO_3 = 137 + 12 + 48 = 197 \ g/mol

4. Determine the number of moles of BaCO_3 in 9.85 g:

\text{Number of moles} = \frac{9.85 \, \text{g}}{197 \, \text{g/mol}} \approx 0.05 \, \text{moles}

5. Since 1 mole of BaCO_3 gives 1 mole of CO_2, 0.05 moles of BaCO_3 will give 0.05 moles of CO_2.
6. The volume of a gas at STP (Standard Temperature and Pressure) is 22.4 liters per mole. So, the volume of CO_2 produced will be:

0.05 \, \text{moles} \times 22.4 \, \text{lit/mole} = 1.12 \, \text{liters}

Thus, the volume of CO_2 obtained by the complete decomposition of 9.85 \, \text{gm} \, BaCO_3 is 1.12 \, \liters.

The correct answer is 1.12 \, \liters.