2CO(g)+O_(2)(g)arrow 2CO_(2)(g) he formation of CO_(2) from CO and oxygen, how many Liters of CO_(2) are produced by the reaction of 8 mols o O_(2) with an excess of carbon monoxide? 0.178 L 358.4 L iters 704 grams 0.36 grams

To solve this problem, we need to use the stoichiometry of the balanced chemical equation and the ideal gas law.<br /><br />Given information:<br />- Balanced chemical equation: $2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g)$<br />- 8 moles of $O_{2}$ are reacted with an excess of carbon monoxide.<br /><br />Step 1: Determine the moles of $CO_{2}$ produced.<br />According to the balanced equation, 1 mole of $O_{2}$ produces 2 moles of $CO_{2}$.<br />Moles of $CO_{2}$ produced = 8 moles of $O_{2} \times \frac{2 moles of $CO_{2}}{1 mole of $O_{2}} = 16 moles of $CO_{2}$<br /><br />Step 2: Calculate the volume of $CO_{2}$ produced using the ideal gas law.<br />At standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 liters.<br />Volume of $CO_{2}$ produced = 16 moles of $CO_{2} \times 22.4 L/mol = 358.4 liters<br /><br />Therefore, the correct answer is:<br />358.4 Liters