Assuming fully decomposed, the volume of $CO_2$ released at STP on heating 9.85 g of $BaCO_3$ (at. mass of Ba = 137) will be

Question

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is:
(Given atomic masses of A=64; B=40; C=32 u)

Answer 1

To find the volume of CO_2 released at STP from heating BaCO_3, we start by analyzing the decomposition reaction:

The chemical reaction for the decomposition of barium carbonate is:

BaCO_3 (s) \rightarrow BaO (s) + CO_2 (g)

From the equation, we see that 1 mole of BaCO_3 releases 1 mole of CO_2.

Let's calculate the molar mass of BaCO_3:

Barium (Ba): 137 \, g/mol
Carbon (C): 12 \, g/mol
Oxygen (O₃): 3 \times 16 \, g/mol = 48 \, g/mol
Total molar mass of BaCO_3: 137 + 12 + 48 = 197 \, g/mol

Now calculate the number of moles of BaCO_3 in 9.85 g:

\text{Moles of } BaCO_3 = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} = \frac{9.85}{197}

Evaluating the above gives:

\text{Moles of } BaCO_3 = \frac{9.85}{197} \approx 0.05 \, \text{moles}

Since 1 mole of BaCO_3 gives 1 mole of CO_2, 0.05 moles of BaCO_3 will produce 0.05 moles of CO_2.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Therefore,

\text{Volume of } CO_2 = 0.05 \times 22.4 \, \text{L} = 1.12 \, \text{L}

Therefore, the volume of CO_2 released at STP is 1.12 L.

Answer 2