can be produced when 1.84 moles of O_(2) reacts with excess iron (Fe)? 4Fe+3O_(2)arrow 2Fe_(2)O_(3) 392g 6.128 39.3g 1968 Multiple Choice 10 points

To determine the amount of product formed when 1.84 moles of $O_{2}$ reacts with excess iron (Fe), we need to use the balanced chemical equation:<br /><br />$4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3}$<br /><br />From the equation, we can see that 3 moles of $O_{2}$ produce 2 moles of $Fe_{2}O_{3}$. Therefore, we can set up a proportion to find the amount of $Fe_{2}O_{3}$ produced from 1.84 moles of $O_{2}$:<br /><br />$\frac{3 \text{ moles of } O_{2}}{2 \text{ moles of } Fe_{2}O_{3}} = \frac{1.84 \text{ moles of } O_{2}}{x \text{ moles of } Fe_{2}O_{3}}$<br /><br />Cross-multiplying and solving for $x$, we get:<br /><br />$3x = 2 \times 1.84$<br /><br />$x = \frac{2 \times 1.84}{3}$<br /><br />$x = 1.2267$ moles of $Fe_{2}O_{3}$<br /><br />Therefore, 1.2267 moles of $Fe_{2}O_{3}$ will be produced when 1.84 moles of $O_{2}$ reacts with excess iron (Fe).