Based on the following equation: 6HCL(aq)+ 2AL(s)> 2ALCL3(aq) + 3H2 (g) What mass of aluminum is needed to produce 500 mL of hydrogen gas at 25 degrees Celcius and 100 kPa?

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Md Ariful Islam · Answer 1 · Apr 23, 2024 13:27:55

P = 100 kPa

T = 25^oC + 273 = 298 K

V = 500 mL = 0.5 L

R = 8.314 kPa L K^-1 mol^-1

Solution:

The balanced chemical equation:

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

According to the chemical equation: n(Al)/2 = n(H₂)/3

The ideal gas law can be used to calculate the amount of H₂:

PV = nRT

n= PV/RT

Thus:

n(H₂) = (100 kPa × 0.5 L) / (8.314 kPa L K^-1 mol^-1 × 298 K) = 0.02 mol

n(Al)/2 = n(H₂)/3

n(Al) = 2 × n(H₂) / 3 = 2 × (0.02 mol) / 3 = 0.0133 mol

Moles of Al = Mass of Al / Molar mass of Al

The molar mass of Al is 27 g/mol

Thus:

Mass of Al = n(Al) × M(Al) = (0.0133 mol) × (27 g/mol) = 0.3591 g = 0.36 g

Answer: 0.36 g of aluminum (Al) is needed.

Based on the following equation: 6HCL(aq)+ 2AL(s)&gt; 2ALCL3(aq) + 3H2 (g) What mass of aluminum is needed to produce 500 mL of hydrogen gas at 25 degrees Celcius and 100 kPa?