We start with the balanced equation.

#"2Al + 3H"_2"SO"_4 → "3H"_2 + "Al"_2("SO"_4)_3#

**Step 1.** Calculate the moles of #"Al"#.

#15.0 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "0.5560 mol Al"#

**Step 2.** Calculate the moles of #"H"_2#.

The balanced equation tells us that 3 mol of #"H"_2# are formed for every 2 mol of #"Al"#. So,

#0.5560 cancel("kmol Na") × ("3 mol Cl"_2)/(2 cancel("mol Al")) = "0.8340 mol H"_2#

**Step3.** To calculate the volume, we use the **Ideal Gas Law**:

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

We can rearrange this to get

#V = (nRT)/p#

In this problem,

#n = "0.8340 mol"#

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

#T = "(22.5 + 273.15) K" = "295.65 K"#

#p = 18.4 color(red)(cancel(color(black)("psi"))) × "1 atm"/(14.7 color(red)(cancel(color(black)("psi")))) = "1.252 atm"#

∴ #V = (0.8340 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 295.65 color(red)(cancel(color(black)("K"))))/(1.252 color(red)(cancel(color(black)("atm")))) = "16.2 L"#