Question M.15 Part B & C (Sixth Edition)

[Steve Magana 2I]

Question: Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255 g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300. g of aluminum chloride has been produced. (a) Write the balanced equation for the reaction. (b) What mass of aluminum chloride can be produced by
these reactants? (c) What is the percentage yield of aluminum chloride?

I know the balanced equation is: 2 Al(s) + 3 Cl2(g) ----> 2 AlCl3 (s) and that the limiting reactant is Al, I just do not know what to do after this. Thank you!

[Heesu_Kim_1F]

Since you know that the limiting reactant is Al, you use the given information about aluminum and do stoichiometry to find the mass of aluminum chloride. First, convert 255g of Al to moles using its molar mass. Then use the balanced equation to convert moles of Al to moles of AlCl3. Then use the molar mass of AlCl3 to convert to grams of AlCl3. That will give you the answer to part (b). For part (c), plug in the answer you got on part (b) as theoretical yield and use the given information 300.g AlCl3 as actual yield. After, utilize the percent yield equation. Hope this helps!

[Steve Magana 2I]

Oops actually Cl2 is the limiting reactant, I accidentally mixed up the results I had calculated with Al.