The ratio of H_(2) gas to O_(2) gas will always remain 2:1 and the ratio of H_(2) gas to H_(2)O gas will remain 1:1 so if the gas volumes are measured at the same temperature and pressure: 44.8 liters of H2+ - liters of O_(2)arrow liters of H_(2)O H_(2)+square

To solve this problem, we need to use the stoichiometry of the chemical reaction between hydrogen gas (\(H_2\)) and oxygen gas (\(O_2\)) to form water vapor (\(H_2O\)). The balanced chemical equation for this reaction is:<br /><br />\[ 2H_2 + O_2 \rightarrow 2H_2O \]<br /><br />From this equation, we can see that:<br /><br />- 2 volumes of \(H_2\) react with 1 volume of \(O_2\) to produce 2 volumes of \(H_2O\).<br /><br />Given that you have 44.8 liters of \(H_2\), let's determine the volumes of \(O_2\) and \(H_2O\) involved in the reaction.<br /><br />1. **Volume of \(O_2\):**<br /><br /> Since the ratio of \(H_2\) to \(O_2\) is 2:1, for every 2 volumes of \(H_2\), you need 1 volume of \(O_2\). Therefore, if you have 44.8 liters of \(H_2\):<br /><br /> \[<br /> \text{Volume of } O_2 = \frac{44.8 \text{ liters of } H_2}{2} = 22.4 \text{ liters of } O_2<br /> \]<br /><br />2. **Volume of \(H_2O\):**<br /><br /> According to the balanced equation, 2 volumes of \(H_2\) produce 2 volumes of \(H_2O\). Thus, the volume of \(H_2O\) produced will be equal to the volume of \(H_2\) used:<br /><br /> \[<br /> \text{Volume of } H_2O = 44.8 \text{ liters}<br /> \]<br /><br />Therefore, the complete reaction is:<br /><br />\[ 44.8 \text{ liters of } H_2 + 22.4 \text{ liters of } O_2 \rightarrow 44.8 \text{ liters of } H_2O \]