We start with the balanced equation.
"2Al + 3H"_2"SO"_4 → "3H"_2 + "Al"_2("SO"_4)_32Al + 3H2SO4→3H2+Al2(SO4)3
Step 1. Calculate the moles of "Al"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"
