Question 0c593

Apr 27, 2017

The volume of hydrogen gas is 16.2 L.

Explanation:

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

Step 1. Calculate the moles of $\text{Al}$.

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

Step 2. Calculate the moles of ${\text{H}}_{2}$.

The balanced equation tells us that 3 mol of ${\text{H}}_{2}$ are formed for every 2 mol of $\text{Al}$. 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:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to get

$V = \frac{n R T}{p}$

In this problem,

$n = \text{0.8340 mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(22.5 + 273.15) K" = "295.65 K}$
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"#