Limiting Reactant Problem

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Kallista McCarty 1C
Posts: 212
Joined: Wed Sep 18, 2019 12:18 am

Limiting Reactant Problem

Postby Kallista McCarty 1C » Tue Oct 08, 2019 1:43 pm

"For the following reaction, how many grams of AlCl3 would be obtained if 5.43g of aluminum and 7.80g of hydrogen chloride (HCl) was used in the reaction?" I know you first balance the equation and I got
2Al + 6 HCl --> 2AlCl3 + 3H2
Can someone show the steps to find how many grams of AlCl3 would be produced?

JasonLiu_2J
Posts: 109
Joined: Sat Aug 24, 2019 12:17 am

Re: Limiting Reactant Problem

Postby JasonLiu_2J » Tue Oct 08, 2019 1:56 pm

The first thing you want to do in this question after balancing the equation is to find the limiting reactant based on your starting amount of each reactant. Convert the amount in grams of each reactant to the number of moles of each reactant that you initially start with. This is done by dividing the masses of each reactant by their respective molar masses. For Al, you would do 5.43 grams of Al divided by 26.98 grams/mol Al, which would give you the moles of Al that you have (which is 0.2013 moles Al). You would then do the same for HCl, by dividing 7.80 grams HCl by its molar mass of 36.46 grams/mol HCl. (This gives 0.214 moles HCl). Now that you have the moles of each reactant, you want to compare that with the moles required in the given reaction to determine which is the limiting reactant. One way to do this is to divide the moles you have by the moles required, and whichever resulting number is lower would be the limiting reactant. Thus, divide 0.2013 moles Al by 2 (from the balanced equation) and 0.214 moles of HCl by 6 (from balanced equation). This would give HCl as the limiting reactant. Another way you could find the limiting reactant is to use the moles of one of the reactants and convert it to the moles of the other reactant using the mole ratios in the balanced equation. If you multiply the moles of Al by 6/2 (ratio of HCl to Al) you will find that 0.2013 mol Al requires 0.6039 moles of HCl to react completely, which is obviously more than the amount you have. Thus, HCl would again be the limiting reactant using this method. Once you have the limiting reactant, convert the number of moles of the limiting reactant to the number of moles of AlCl3 that could be produced if the limiting reactant were to be used up completely. Thus, multiply 0.214 moles of HCl by the mole ratio of Al2Cl3 to HCl (2/6). After you have moles of Al2Cl3, simply multiply the moles by the molar mass of Al2Cl3 to find the amount in grams of Al2Cl3 produced. Remember, the amount of product that can be produced is based on how much limiting reactant you have initially! Hope this helps.

Kallista McCarty 1C
Posts: 212
Joined: Wed Sep 18, 2019 12:18 am

Re: Limiting Reactant Problem

Postby Kallista McCarty 1C » Tue Oct 08, 2019 2:02 pm

Thank you so much!

Jorge Ramirez_4H
Posts: 54
Joined: Wed Nov 21, 2018 12:19 am

Re: Limiting Reactant Problem

Postby Jorge Ramirez_4H » Tue Oct 08, 2019 3:02 pm

Tips for rearranging for these types of problems?

MMckinney_4H
Posts: 61
Joined: Sat Sep 28, 2019 12:16 am

Re: Limiting Reactant Problem

Postby MMckinney_4H » Tue Oct 08, 2019 5:29 pm

Which question was this?

MMckinney_4H
Posts: 61
Joined: Sat Sep 28, 2019 12:16 am

Re: Limiting Reactant Problem

Postby MMckinney_4H » Tue Oct 08, 2019 5:40 pm

This is how I did it step by step.
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