Answer to Question #242805 in General Chemistry for Brodi Levitt

Question

If you had excess aluminum, how many moles of aluminum chloride could be produced from 18.0g of chlorine gas, Cl_2?

Answer 1

2Al (s) + 3Cl2 (g) --> 2AlCl3 (s)

mass of chlorine gas we are given = 18.0g

So amount of aluminum chloride produced will depend on the mass of chlorine.

Atomic mass of Al = 35.45g/mol

moles of Cl = 18g cl * 1mol Al / 35.45g Cl = 0.51 mol of Cl

Next we use the mole ratio of Cl and AlCl3 from the reaction and find the ‘moles of AlCl3 ‘ .

mol ratio of Cl and AlCl3 in the reaction = 3 : 2

So, moles of AlCl3 formed = 2/3 × 0.51

= 0.34mol