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
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Ответ:
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
Ответ:
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)
Ответ:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48