Answer:
3.65 liters is the volume of H₂ produced in the reaction
Explanation:
This is the reaction:
2 Al(s) + 6 HCI (aq) → 2AICI₃ (aq) + 3 H₂ (g)
Let's convert the mass of Al to moles.
2.93 g . 1mol/26.98 g = 0.108 moles
Ratio is 2:3. 2 moles of Al can produce 3 moles of H₂
Then, 0.108 moles of Al would produce (0.108 .3)/ 2 = 0.163 moles of H₂
Let's apply the Ideal Gases Law to find out the produced volume, at STP.
STP is 1 atm as pressure and 273K as T°
1 atm . V = 0.163 mol . 0.082L.atm/mol.K . 273K
V = (0.163 mol . 0.082L.atm/mol.K . 273K) / 1atm
V = 3.65 L
3.65 L of hydrogen gas will be produced.
Given:
Mass of Al = 2.93 g
1 mole of a gas = 22.4 L of volume
To find:
Volume of hydrogen gas produced = ?
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Putting mass of Al as 2.93 g and molar mass as 26.98 g/mol in the formula as follows:
[tex]\text{Moles of Al}=\frac{2.93 \text{g}}{26.98 \text{g/mol}}\\\\\text{Moles of Al}=0.1086\text{ mol}[/tex]
By stoichiometry of the reaction:
If 2 moles of aluminum produces 3 moles of hydrogen gas
So, 0.1086 moles of aluminum will produce = [tex]\frac{3}{2}\times 0.1086=0.1629mol[/tex] of hydrogen gas
At STP conditions:
0.1629 moles of hydrogen gas will have a volume of = [tex]\frac{22.4 L}{1 mol}\times 0.1629mol=3.65L[/tex]
Thus, 3.65 L of hydrogen gas will be produced.
Learn more about STP conditions: https://brainly.com/question/6439270
