47.3 L of Hydrogen gas is produced from 2.11 moles of Aluminum metal.
Explanation:
We have to find the volume of hydrogen produced by writing the balanced equation as,
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
We can use the ideal gas equation at STP,
Temperature = 273.15K
Pressure = 1 atm
Gas constant being 0.08205 L atm K⁻¹ mol⁻¹
Number of moles = 2.11 moles
PV = nRT
Plugin the values as,
[tex]$ V = \frac{nRT}{P}[/tex]
[tex]$V = \frac{2.11 \times 0.08205 \times 273.15}{1}[/tex]
= 47.3 L
So 47.3 L of Hydrogen gas is produced from 2.11 moles of Aluminum metal.
