Question

If you had excess chlorine, how many moles of aluminum chloride could be produced from 28.0 g of aluminum?

Answer 1

Answer:

1.04 moles of AlCl₃

Explanation:

Balanced Equation:  2Al°(s) + 3Cl₂(g) => 2AlCl₃(s)

Approach => convert given data (28 g Al°) to moles. The number of moles  AlCl₃ would equal the number of moles aluminum because coefficients for Aluminum and Aluminum Chloride are equal.

moles Al° = (28g)/(27g/mole ) = 1.04 mole Al°

From balanced equation moles AlCl₃ produced equals moles of Al° produced = 1.04 mole AlCl₃ because coefficients of Al° & AlCl₃.