The reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the hydrogen gas collected occupies 14 L at 300 K and was produced upon reaction of 4.5 mol of Al? Assume the yield of the reaction is 100 percent. Assume the volume, temperature, moles, and presure are actually known to three signficant figures. 1. 1.07 atm 2. 0.0763 atm 3. 0.233 atm 4. 7.91 atm 5. 11.9 atm 6. 5.28 atm

Pressure for H₂ = 11.9 atm

Option 5.

Explanation:

We determine the complete reaction:

2Al(s) +  6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

As we do not know anything about the HCl, we assume that the limiting reactant is the Al and the acid is the excess reagent.

Ratio is 2:3.

2 moles of Al, can produce 3 moles of hydrogen

Therefore 4.5 moles of Al must produce (4.5 . 3) / 2 = 6.75 moles

Now we can apply the Ideal Gases law to find the H₂'s pressure

P . V  = n . R . T → P = (n . R .T) / V

We replace data: (6.75 mol . 0.082L.atm/mol.K . 300K) / 14L

Pressure for H₂ = 11.9 atm

The pressure of H2 is 11.9 atm (option 5)

Explanation:

Step 1: Data given

Volume of hydrogen gas = 14L

Temperature = 300 K

Moles of Al = 4.5 moles

Step 2: The balanced equation

2Al + 6HCl → 2AlCl3 + 3H2

Step 3: Calculate moles of H2

For 2 moles Al we need 6 moles HCl to produce 2 moles AlCl3 and 3 moles H2

For 4.5 moles Al we need 3*4.5 moles HCl = 13.5 moles HCl

We'll produce 4.5 moles AlCl3 and 3/2 * 4.5 = 6.75 moles H2

Step 4: Calculate pressure of H2

p*V = n*R*T

⇒with p = the pressure of H2 = TO BE DETERMINED

⇒with V = the volume of H2 = 14 L

⇒with n = the moles H2 = 6.75 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 300 K

p = (n*R*T) / V

p = (6.75 * 0.08206 * 300 ) / 14

p = 11.9 atm

The pressure of H2 is 11.9 atm (option 5)

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