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Calculate the current in amperes required to produce 18.0g of aluminium in 1.50hrs (al=27.og )

Sagot :

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Answer:

35.7 A

Explanation:

Let's consider the reduction half-reaction of Al³⁺.

Al³⁺ + 3 e⁻ ⇒ Al

We will calculate the charge required to produce 18.0 g of Al using the following conversion factors.

  • 1 mole of Al has a mass of 27.0 g
  • 1 mole of Al is formed upon the circulation of 3 moles of electrons
  • 1 mole of electrons has a charge of 96486 C (Faraday's constant)

[tex]18.0gAl \times \frac{1molAl}{27.0gAl} \times \frac{3mole^{-} }{1molAl} \times \frac{96486C}{1mole^{-} } = 1.93 \times 10^{5} C[/tex]

1.93 × 10⁵ C circulate in 1.50 hours. The intensity is:

[tex]I = \frac{1.93 \times 10^{5} C}{1.50h} \times \frac{1h}{3600s} = 35.7 A[/tex]

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