Question

If 0.1453 g of aluminum metal reacted with excess HCl(aq) in an experiment conducted in Denver, CO, calculate the theoretical volume of H2 that would be collected over water if the temperature of was 17.0 C and the atmospheric pressure was 29.25 inches of Hg. (use textbook R value)

Answer 1

Answer:

0,1966L

Explanation:

The reaction of aluminium with HCl is:

Al(s) + 3HCl(aq) → AlCl₃ + ³/₂H₂(g)

Moles of aluminium that react are:

0,1453g×$\frac{1mol}{26,98g}$= 5,385x10⁻³ moles of Al

As 1 mole of Al produce ³/₂ moles of H₂(g), the moles of H₂ are:

5,385x10⁻³ moles of Al×$\frac{3/2molH_2}{1molAl}$ = 8,078x10⁻³ mol H₂(g)

To find the volume H₂ occupies you need to use:

$V =\frac{nRT}{P}$

Where n are moles (8,078x10⁻³ mol H₂(g))

R is gas constant (0,082atmL/molK)

T is temperature (17,0°C+273,15 = 290,15K)

And P is pressure (29,25 inches of Hg/29,921= 0,9776 atm)

Replacing:

$V = 0,1966L$

I hope it helps!