To determine the volume of [tex]\( \text{CO}_2 \)[/tex] gas produced at standard temperature and pressure (STP) from the decomposition of 3.25 moles of [tex]\( \text{CaCO}_3 \)[/tex], follow these steps:
1. Write the balanced chemical equation:
[tex]\[
\text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g)
\][/tex]
2. Identify the stoichiometric relationship:
From the balanced equation, 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
3. Determine the amount of [tex]\( \text{CO}_2 \)[/tex] produced:
Given that 3.25 moles of [tex]\( \text{CaCO}_3 \)[/tex] decompose, this will produce 3.25 moles of [tex]\( \text{CO}_2 \)[/tex], as per the 1:1 molar ratio.
4. Use the molar volume of a gas at STP:
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.
5. Calculate the volume of [tex]\( \text{CO}_2 \)[/tex] gas produced:
[tex]\[
\text{Volume of } \text{CO}_2 = \text{moles of } \text{CO}_2 \times \text{volume per mole at STP}
\][/tex]
Substituting the values:
[tex]\[
\text{Volume of } \text{CO}_2 = 3.25 \text{ moles} \times 22.4 \text{ L/mole}
\][/tex]
6. Compute the result:
[tex]\[
\text{Volume of } \text{CO}_2 = 3.25 \times 22.4 = 72.8 \text{ liters}
\][/tex]
Therefore, the volume of [tex]\( \text{CO}_2 \)[/tex] gas produced at STP is [tex]\( 72.8 \)[/tex] liters.
