Question

Fan, Medal, and Testimonial.


How many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is 2.70 g/mL? Show all steps of your calculation as well as the final answer.

AlCl3 → Al + Cl2

@bbypanda16

Answer 1

You must start with a balanced chemical equation:
2AlCl3 → 2Al + 3Cl2
???g ......78.3 mL

Use the volume and density to get the mass, and then the number of moles of Al, and then the moles of Al2O3 (it will be the same, since the mole ratio is 1:1) and finally convert the moles of AlCl3 to grams of AlCl3.

78.3 mL x (2.70g Al / 1 mL) x (1 mol Al / 27.0g Al) x (1 mol AlCl3 / 1 mol Al) x (133.5 g AlCl3 / 1 mol AlCl3) = 1045 g AlCl3