Aluminum reacts with chlorine gas toform aluminum chloride via the following rea

Question

Aluminum reacts with chlorine gas toform aluminum chloride via the following reaction: rm 2Al{ it (s)}+3Cl 2{ it (g)} rightarrow2AlCl 3{ it(s)} You are given 28.0 g of aluminum and 33.0 g of chlorine gas. Part A If you had excess chlorine, how many moles ofof aluminum chloride could be produced from 28.0 g of aluminum? Express your answer numerically inmoles. 1.04 rm mol~AlCl 3 Part B If you had excess aluminum, how many moles ofaluminum chloride could be produced from 33.0 g of chlorine gas, rm Cl 2? Express your answer numerically inmoles. 0.310 rm mol~AlCl 3 By comparing your answers for Parts A and B, you candetermine which reactant is limiting. Keep in mind that thelimiting reactant is the one that produces the lesser amount ofproduct. PartC Aluminum reacts with chlorine gas to formaluminum chloride via the following reaction: rm 2Al{ it (s)}+3Cl 2{ it (g)} rightarrow2AlCl 3{ it(s)} What is the maximum mass of aluminum chloride that can be formedwhen reacting 28.0 g of aluminum with 33.0 g of chlorine? Express your answer numerically ingrams. I understand I need to add Al+Cl,just not sure how to calculate it right.

Explanation / Answer

Since you can only make 0.310 moles of product with the givenamount of chlorine and excess aluminum, and you can make 1.04 molesof product with the given amount of aluminum and excess chlorine,the limiting reactant is chlorine. This also means that the maximummoles of product that can be made is 0.310 moles. Once you use upall the chlorine you are given, you no longer have any moreingredients to make the product, and so the production stops at0.310 moles. To answer the question, simply convert 0.310 moles to grams ofaluminum chloride. 0.310 moles AlCl3 x 133.34 g/mol AlCl3 = 41.336 gramsAlCl3

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