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### M. 15 Fundamentals

Posted: **Thu Oct 05, 2017 8:24 pm**

by **alyssawhite1L**

Here is the problem:

Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300g 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?

In order to begin this problem, you must balance the equation, but I cant even remember how to write the formula for the molecules! How do you derive the formula for aluminum chloride if it is not given?

### Re: M. 15 Fundamentals

Posted: **Thu Oct 05, 2017 9:18 pm**

by **Vasiliki G Dis1C**

To derive the formula for aluminum chloride, you can use the oxidation numbers for the elements involved. Aluminum has an oxidation number of 3 so you would write it as Al and then +3 as an exponent. Cl has an oxidation number of -1 so you would write it as Cl with -1 as the exponent. Now you have Al+3Cl=1, and you can take these exponents and criss cross them (if that makes sense). You would put the 1 as a subscript for Al and the 3 as a subscript for Cl, making the overall formula AlCl3. Hope this helps!!

### Re: M. 15 Fundamentals

Posted: **Thu Oct 05, 2017 9:20 pm**

by **Jessica Beroukhim 3K**

It looks like your question is about how to write chemical formulas when it's not given in problems like this. I don't think we're going to be asked to do that on tomorrow's test but I'll answer you anyways in a very simple way (just enough to get by).

Ions of elements in group 17 (which CL is in) have a 1- charge. Ions of elements in group 13 (where Al is) have a 3+ charge. You can remember these just by counting the groups - (1 = 1+, 2 = 2+, 13 = 3+, 15 = 3-, 16 = 2-, 17 = 1-).

When the ions go together they have to balance and become neutrally charged. You need 3 Cl ions of 1- charge to balance 1 Al ion of 3+ charge. Therefore the formula is AlCl_{3} You can also do this by making the charge of one of them the number of the other (you might remember this from high school as the crossing method).

Hope this helps!

### Limiting Reactant

Posted: **Thu Oct 05, 2017 9:29 pm**

by **Gabriela Carrillo 1B**

After figuring out the limiting reactant, how do I calculate how much product is produced and how much excess is produced? Do I multiply the amount of moles available of the limiting reactant by the ratio of moles in the equation???

### Re: M. 15 Fundamentals

Posted: **Thu Oct 05, 2017 10:42 pm**

by **Jana Sun 1I**

You're on the right track! After you figure out the limiting reactant, you can start with the amount of that limiting reactant you're given, whether that's moles or grams, then use dimensional analysis with molar masses and molar ratios to determine the amount of product produced.

In this problem, if we already know that Cl2 is the limiting reactant and you know that we start with 535g of Cl2, we can convert this amount to moles Cl2 by dividing it by the molar mass of Cl2 (70.906g/mol). Then, we'll need to use our molar ratio from our balanced chemical equation to convert moles Cl2 to moles product. 3 moles of Cl2 are needed to create 2 moles of AlCl3 so we multiply our moles Cl2 by (2molAlCl3 / 3molCl2). Then we're left with our moles of product and you can convert moles AlCl3 into grams AlCl3 by multiplying it by the molar mass of AlCl3 (133.321g/mol)!

Hope this makes sense!