Module 3: Limiting Reactant Calculations Problem 21  [ENDORSED]

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Sara Veerman-1H
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Module 3: Limiting Reactant Calculations Problem 21

Postby Sara Veerman-1H » Fri Apr 06, 2018 4:52 pm

After doing this problem several times over, I never end up with any of the given answers.

The problem gives 0.750 g C6H9Cl3, 1.000 kg AgNO3, and their molar masses: 187.50 g/mol and 169.88 g/mol respectively.
Chemical Formula: C6H9Cl3 + 3(AgNO3) ---> AgCl + C6H9(NO3)3.

I found that there are 4x10^3 mol C6H9Cl3 and 5.885 mol AgNO3 and that C6H9Cl3 is the limiting reagent.
However, when I calculate for the mass of AgCl, I get 0.573 g.

The possible answers are 1.00 g, 1.14 g, 1.50 g, and 1.72 g AgCl.

I don't know what I'm doing wrong...

Mei Blundell_1J
Posts: 32
Joined: Fri Apr 06, 2018 11:01 am
Been upvoted: 1 time

Re: Module 3: Limiting Reactant Calculations Problem 21  [ENDORSED]

Postby Mei Blundell_1J » Fri Apr 06, 2018 5:17 pm

Hi, I think you forgot to balance the equation. With the balanced equation, the limiting reactant will be the same as what you found, but you will have a different molar ratio to work with to find the mass of AgCl.

Sara Veerman-1H
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Re: Module 3: Limiting Reactant Calculations Problem 21

Postby Sara Veerman-1H » Fri Apr 06, 2018 8:00 pm

Ohh thanks! I missed that


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