In this question, the first step is to set up the properly balanced equation:
2 AlCl3 + 9 O2 -> 2 Al(ClO3)3
We have:
20.6 grams of AlCl3
We need to find the number of moles of AlCl3, and check the number of moles of O2 using the molar ratio concept, which according to the reaction, the molar ratio will be 2 moles of AlCl3 for every 9 moles of O2.
Using the mass given in the question and the molar mass of AlCl3, 133.34g/mol, we have:
133.34g = 1 mol
20.6g = x moles
x = 0.154 moles of AlCl3
Now using the molar ratio, we will have:
2 AlCl3 = 9 O2
0.154 AlCl3 = x O2
x = 0.693 moles of O2 are needed to react with 20.6 grams of AlCl3