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 hydrogen were required to form 3.00 moles of water? 0.0469 g 1.49 g 0.672 g 6.05 g

To determine how many grams of hydrogen were required to form 3.00 moles of water, we need to follow these steps:<br /><br />1. **Understand the Reaction**: The balanced chemical equation is:<br /> \[<br /> 2H_2 + O_2 \rightarrow 2H_2O<br /> \]<br /> This means 2 moles of \( H_2 \) react with 1 mole of \( O_2 \) to produce 2 moles of \( H_2O \).<br /><br />2. **Determine Moles of Hydrogen Used**: According to the stoichiometry of the reaction, forming 2 moles of water requires 2 moles of hydrogen. Therefore, forming 3 moles of water would require:<br /> \[<br /> \frac{3 \text{ moles } H_2O}{2 \text{ moles } H_2O} \times 2 \text{ moles } H_2 = 3 \text{ moles } H_2<br /> \]<br /><br />3. **Calculate Mass of Hydrogen**: The molar mass of hydrogen (\( H_2 \)) is approximately 2.02 g/mol. Therefore, the mass of 3 moles of hydrogen is:<br /> \[<br /> 3 \text{ moles } \times 2.02 \text{ g/mol} = 6.06 \text{ g}<br /> \]<br /><br />Thus, the correct answer is approximately 6.05 g (considering rounding differences). Therefore, the closest option is:<br /><br />6.05 g