Limiting Reactant Calculation post-module question

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Rebecca Epner 4A
Posts: 53
Joined: Sat Aug 17, 2019 12:18 am

Limiting Reactant Calculation post-module question

Postby Rebecca Epner 4A » Mon Sep 30, 2019 9:05 pm

Can someone explain the following problem?

According to the following equation, 0.750 g of C6H9Cl3 is mixed with 1.000 kg of AgNO3 in a flask of water. A white solid, AgCl, completely precipitates out. What is the mass of AgCl produced?
C6H9Cl3 + 3AgNO3 ---> AgCl + C6H9(NO3)3

I found the molar masses of both reactants and determined the limiting reagent. There is .004 mol of C6H9Cl3 so I multiplied that by 3 to get .012. What is the next step?

Abigail Sanders 1E
Posts: 112
Joined: Wed Sep 11, 2019 12:16 am

Re: Limiting Reactant Calculation post-module question

Postby Abigail Sanders 1E » Mon Sep 30, 2019 9:08 pm

The next step would be to convert the 0.012 moles to grams. This should give you the correct answer on the module since the question is asking for grams of the product.

Jesse H 2L
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Joined: Fri Aug 09, 2019 12:17 am
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Re: Limiting Reactant Calculation post-module question

Postby Jesse H 2L » Tue Oct 01, 2019 12:28 am

convert to grams

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: Limiting Reactant Calculation post-module question

Postby Justin Vayakone 1C » Tue Oct 01, 2019 12:38 am

Since C6H9Cl3 is the limiting reactant and AgCl is the product we want to calculate, we need to focus on the molar ratio between them, which is 1 to 1. Because the molar ratio is 1 to 1 (meaning they have the same coefficients in the balanced equation), 0.004 mol C6H9Cl3 produces 0.004 mol AgCl. Then to convert this into mass, we multiply the moles by the molar mass of AgCl (143.32g/mol), which gives us a mass of 0.573 grams AgCl.

Ada Chung 1C
Posts: 55
Joined: Sat Aug 17, 2019 12:15 am

Re: Limiting Reactant Calculation post-module question

Postby Ada Chung 1C » Tue Oct 01, 2019 10:35 am

Since you already calculated the number of moles of C6H9Cl3 and we can tell that C6H9Cl3 is the limiting reactant we would need to figure out the amount of AgCl.
Steps I would take:
1) Convert grams to moles of C6H9Cl3 (which you already did)
2) Figure out the limiting reaction (in our case it is C6H9Cl3)
3) Making sure the chemical equation is balanced find the ratio between C7H9Cl3 to AgCl (our product which is on the righthand side)
4) Since the ratio is 1:1 there would also be 0.004 mol AgCl
5) Convert moles of AgCl to grams of AgCl to find mass (0.004 x molar mass of AgCl) gives us a product of 0.57 grams AgCl


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