For 2H_(2)+O_(2)arrow 2H_(2)O 4 moles of H_(2) will react with square moles of O_(2) to produce square moles of H_(2)O

To solve this problem, we need to use the stoichiometry of the balanced chemical equation:<br /><br />\[ 2H_{2} + O_{2} \rightarrow 2H_{2}O \]<br /><br />From the equation, we can see that 2 moles of \( H_{2} \) react with 1 mole of \( O_{2} \) to produce 2 moles of \( H_{2}O \).<br /><br />Given that you have 4 moles of \( H_{2} \), we can set up a proportion based on the stoichiometry:<br /><br />- For every 2 moles of \( H_{2} \), 1 mole of \( O_{2} \) is required.<br /><br />Therefore, for 4 moles of \( H_{2} \):<br /><br />\[<br />\frac{4 \text{ moles } H_{2}}{2 \text{ moles } H_{2}} = \frac{x \text{ moles } O_{2}}{1 \text{ mole } O_{2}}<br />\]<br /><br />Solving for \( x \):<br /><br />\[<br />x = \frac{4}{2} \times 1 = 2 \text{ moles of } O_{2}<br />\]<br /><br />Now, using the same stoichiometric ratio, 4 moles of \( H_{2} \) will produce:<br /><br />- For every 2 moles of \( H_{2} \), 2 moles of \( H_{2}O \) are produced.<br /><br />So, for 4 moles of \( H_{2} \):<br /><br />\[<br />\frac{4 \text{ moles } H_{2}}{2 \text{ moles } H_{2}} = \frac{y \text{ moles } H_{2}O}{2 \text{ moles } H_{2}O}<br />\]<br /><br />Solving for \( y \):<br /><br />\[<br />y = \frac{4}{2} \times 2 = 4 \text{ moles of } H_{2}O<br />\]<br /><br />Thus, 4 moles of \( H_{2} \) will react with **2 moles** of \( O_{2} \) to produce **4 moles** of \( H_{2}O \).