The volume of $CO_2$ obtained by the complete decomposition of $9.85 ext{ g}$ of $BaCO_3$ is:

Question

The volume of $CO_2$ obtained by the complete decomposition of $9.85	ext{ g}$ of $BaCO_3$ is:

Accepted Answer

1. **Write the balanced chemical equation:**
   $$ BaCO_3(s) \xrightarrow{\Delta} BaO(s) + CO_2(g) $$
2. **Calculate the Molar Mass of $BaCO_3$:**
   Atomic masses: $Ba = 137$, $C = 12$, $O = 16$.
   $$ M = 137 + 12 + (3 	imes 16) = 137 + 12 + 48 = 197	ext{ g/mol} $$
3. **Calculate moles of $BaCO_3$:**
   $$ n = \frac{	ext{Given Mass}}{	ext{Molar Mass}} = \frac{9.85	ext{ g}}{197	ext{ g/mol}} = 0.05	ext{ mol} $$
4. **Apply Stoichiometry:**
   From the equation, $1	ext{ mol}$ of $BaCO_3$ produces $1	ext{ mol}$ of $CO_2$. Therefore, $0.05	ext{ mol}$ of $BaCO_3$ will produce $0.05	ext{ mol}$ of $CO_2$.
5. **Calculate Volume at STP:**
   At STP, $1	ext{ mol}$ of any ideal gas occupies $22.4	ext{ L}$.
   $$ V = n 	imes 22.4	ext{ L} = 0.05 	imes 22.4 = 1.12	ext{ L} $$

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