8. How many liters of CO_(2) are produced from 5.00 moles of CaCO_(3) at STP? CaCO_(3)arrow CaO+CO_(2) 4.48 liters 5.00 liters 22.4 liters 112 liters

To solve this problem, we need to use stoichiometry and the ideal gas law concept at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies **22.4 liters**.<br /><br />### Step-by-Step Solution:<br /><br />1. **Write the balanced chemical equation:**<br /> \[<br /> CaCO_3 \rightarrow CaO + CO_2<br /> \]<br /> From the equation, we see that **1 mole of \(CaCO_3\)** produces **1 mole of \(CO_2\)**.<br /><br />2. **Determine the moles of \(CO_2\) produced:**<br /> Since the reaction is a 1:1 molar ratio, if we start with **5.00 moles of \(CaCO_3\)**, it will produce **5.00 moles of \(CO_2\)**.<br /><br />3. **Convert moles of \(CO_2\) to volume at STP:**<br /> At STP, 1 mole of gas occupies **22.4 liters**. Therefore:<br /> \[<br /> \text{Volume of } CO_2 = \text{moles of } CO_2 \times 22.4 \, \text{L/mole}<br /> \]<br /> Substituting the values:<br /> \[<br /> \text{Volume of } CO_2 = 5.00 \, \text{moles} \times 22.4 \, \text{L/mole} = 112 \, \text{liters}.<br /> \]<br /><br />### Final Answer:<br />The volume of \(CO_2\) produced is **112 liters**.<br /><br />Correct option: **112 liters**