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

To determine the volume of \( CO_2 \) produced from 5.00 moles of \( CaCO_3 \) at STP, we need to use the stoichiometry of the reaction and the molar volume of a gas at STP.<br /><br />The balanced chemical equation is:<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 />At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.<br /><br />Given:<br />- 5.00 moles of \( CaCO_3 \)<br /><br />Since the reaction produces 1 mole of \( CO_2 \) for every 1 mole of \( CaCO_3 \):<br /><br />\[ 5.00 \text{ moles of } CaCO_3 \times \frac{1 \text{ mole of } CO_2}{1 \text{ mole of } CaCO_3} = 5.00 \text{ moles of } CO_2 \]<br /><br />Now, we convert moles of \( CO_2 \) to liters at STP:<br /><br />\[ 5.00 \text{ moles of } CO_2 \times 22.4 \text{ liters/mole} = 112 \text{ liters} \]<br /><br />Therefore, the volume of \( CO_2 \) produced from 5.00 moles of \( CaCO_3 \) at STP is 112 liters.<br /><br />So, the correct answer is:<br />\[ \boxed{112 \text{ liters}} \]