What is the maximum mass of aluminum chloride that can be formed when reacting 1

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First find the number of moles of each species you have using n = m/M: Aluminum: n = 15.0 / 26.98 = 0.55 moles of Al 2Al(s) + 3Cl_2(g) -----> 2AlCl_3(s) Chlorine: n = 20.0 / (35.453 x 2) = 0.282 moles of Cl2 It is obvious that Cl2 is the limiting reagent so this will determine the amount of aluminum chloride that will be formed. By refering to the equation 3 moles of chlorine gas makes 2 moles of AlCl3 so: Moles of AlCl3 formed = 2/3 x 0.282 moles = 0.188 moles Using n = m/M 0.188 = m / (26.98 + 35.453 x 3) m = 25.12 grams So 25.12 grams of AlCl3 will be formed.

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