Problem From Limiting Reactant Module

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Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am

Problem From Limiting Reactant Module

Postby Jocelyn Thorp 1A » Thu Oct 03, 2019 3:18 pm

Hey all, I was hoping for help on one of the problems from the module that I simply wasn't able to get even a remotely close answer for. Any explanation would be very helpful. The problem is as follows:

22. 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.

Molar Mass: C6H9Cl3 (187.50 g/mol), 3AgNO3 (169.88 g/mol), AgCl (143.32 g/mol)

Daria Azizad 1K
Posts: 116
Joined: Thu Jul 25, 2019 12:15 am

Re: Problem From Limiting Reactant Module

Postby Daria Azizad 1K » Thu Oct 03, 2019 3:45 pm

Hi, to solve this problem, first the equation needs to be fully balanced. There's already a stoichiometric coefficient in front of AgNO3, however there also needs to be a 3 in front of AgCl. So, the resulting equation should be:

C6H9Cl3 + 3AgNO3 ---> 3AgCl + C6H9(NO3)3

Then, find the molar masses of each reactant.

(0.750g C6H9Cl3)/(187.50g/mol)= 0.00400mol
(1000g AgCl) / (169.88g/mol) = 5.89mol

Since the ratio is 1:3, we need 3x the amount of AgNO3 than C6H9Cl3.

0.00400*3=0.0120. We have more moles of AgNO3 than that, so AgNO3 is INXS and C6H9Cl3 is the LR.

Now, convert moles of C6H9Cl3 to grams AgCl using the ratio from the balanced eqn and AgCl's molar mass.

(0.00400mol C6H9Cl3) *(3 mol AgCl/1 mol C6H9Cl3) *(143.32g/mol)=1.72g AgCl

Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am

Re: Problem From Limiting Reactant Module

Postby Jocelyn Thorp 1A » Fri Oct 04, 2019 9:49 am

Daria Azizad 1K wrote:Hi, to solve this problem, first the equation needs to be fully balanced. There's already a stoichiometric coefficient in front of AgNO3, however there also needs to be a 3 in front of AgCl. So, the resulting equation should be:

C6H9Cl3 + 3AgNO3 ---> 3AgCl + C6H9(NO3)3

Then, find the molar masses of each reactant.

(0.750g C6H9Cl3)/(187.50g/mol)= 0.00400mol
(1000g AgCl) / (169.88g/mol) = 5.89mol

Since the ratio is 1:3, we need 3x the amount of AgNO3 than C6H9Cl3.

0.00400*3=0.0120. We have more moles of AgNO3 than that, so AgNO3 is INXS and C6H9Cl3 is the LR.

Now, convert moles of C6H9Cl3 to grams AgCl using the ratio from the balanced eqn and AgCl's molar mass.

(0.00400mol C6H9Cl3) *(3 mol AgCl/1 mol C6H9Cl3) *(143.32g/mol)=1.72g AgCl


Thank you so much!! I assumed the equation was balance (because there was already a stoichiometric coefficient) but that's what threw the whole answer off.


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