If you had excess chlorine, how many moles of of aluminum chloride could be produced from 23.0 gg of aluminum? Express your answer to three significant figures and include the appropriate units.

0.582 moles of aluminum chloride are produced

Explanation:

We determine the reaction to begin the excersise:

2Al (s) + 3Cl₂(g) → 2AlCl₃

As the chlorine is in excess, Al must be the limiting reactant. And the ratio is 2:2, so the moles I used from Al will be the same moles of chloride produced.

We convert the mass from Al to moles:

23 g . 1 mol / 26.98g = 0.852 moles

In conclussion we produce 0.582 moles of aluminum chloride

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