**Step 1. Write the balanced chemical equation**

#M_text(r):color(white)(mmmmmml)18.02#

#color(white)(mm)"2H"_2 + "O"_2 → "2H"_2"O"#

**Step 2. Calculate the moles of hydrogen**

We can use the **Ideal Gas Law** to calculate the number of moles.

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this formula to get

#n = (pV)/(RT)#

In this problem,

#p = 801 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "1.054 atm"#

#V = "13.74 L"#

#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#

#T = "(30.0 + 273.15) K" = "303.15 K"#

#n = (1.054 cancel("atm") × 13.74 cancel("L"))/("0.082 06" cancel("L·atm·K"^"-1")"mol"^"-1" × 303.15 cancel("K")) = "0.5821 mol"#

**Step 3. Calculate the moles of water formed**

#"Moles of H"_2"O" = 0.5821 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.5821 mol H"_2"O"#

**Step 4. Calculate the mass of water formed**

#"Mass of H"_2"O" = 0.5821 cancel("mol H"_2"O") × ("18.02 g H"_2"O")/(1 cancel("mol H"_2"O")) = "10.5 g H"_2"O"#