Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) + 3Cl2(g) +2AICI3(s)
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2Al(s) + 3Cl2(g) +2AICI3(s)
You are given 12.0 g of aluminum and 170 g of chlorine gas.
Part A
If you had excess chlorine, how many moles of aluminum chloride could be produced from 12.0 g of aluminum?
Homework Answers
Molar mass of Al = 26.98 g/mol
mass of Al = 12 g
mol of Al = (mass)/(molar mass)
= 12/26.98
= 0.4448 mol
According to balanced equation
mol of AlCl3 formed = moles of Al
= 0.4448 mol
Answer: 0.445 mol
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