When 3.00 moles of hydrogen molecules and 1.50 moles of oxygen molecules react, they form 3.00 moles of water according to the reaction below. 2H_(2)+O_(2)arrow 2H_(2)O How many grams of oxygen were required? 48.09 0.0469 g 0.0208 g 21.3 g

To determine how many grams of oxygen were required, we need to use stoichiometry based on the balanced chemical equation:<br /><br />\[ 2H_2 + O_2 \rightarrow 2H_2O \]<br /><br />From the equation, 1 mole of \( O_2 \) reacts with 2 moles of \( H_2 \) to produce 2 moles of \( H_2O \).<br /><br />Given:<br />- 3.00 moles of \( H_2 \)<br />- 1.50 moles of \( O_2 \)<br /><br />According to the stoichiometry of the reaction, 1.50 moles of \( O_2 \) are needed to react with 3.00 moles of \( H_2 \), which is consistent with the given amounts.<br /><br />Now, calculate the mass of 1.50 moles of \( O_2 \):<br /><br />The molar mass of \( O_2 \) is approximately 32.00 g/mol.<br /><br />\[ \text{Mass of } O_2 = \text{moles} \times \text{molar mass} = 1.50 \, \text{moles} \times 32.00 \, \text{g/mol} = 48.00 \, \text{g} \]<br /><br />Therefore, the correct answer is approximately 48.09 g (considering rounding differences).