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

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 g1 mole of Al is formed upon the circulation of 3 moles of electrons1 mole of electrons has a charge of 96486 C (Faraday's constant)

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

The formula for density = mass/volume.
We know the density (21.45 g/cm³)
We know the volume (80.00 cm³) of the sample
We don't know the mass, so first - solve the equation for mass:

D=m/v  [multiply v to both sides to isolate m]
Dv=m

Now, insert the known value and units...make sure the units cancel properly:
21.45 g/cm³ x 80.00 cm³ = 1716 g