Mass Yield of Product

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EMurphy_2L
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

Mass Yield of Product

Postby EMurphy_2L » Sat Sep 28, 2019 7:08 pm

Question: 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)

My issue: I converted each reactant to moles and found C6H9Cl3 is the limiting reactant (because there are only .004 moles of it in the reaction). This should yield .004 moles of AgCl, right? So the question is asking what the mass is however my calculated mass is not an answer choice. I'm not sure if I have a concept issue or a computational issue. Any help is appreciated!

Kellylin_4D
Posts: 66
Joined: Sat Aug 24, 2019 12:17 am

Re: Mass Yield of Product

Postby Kellylin_4D » Sat Sep 28, 2019 7:26 pm

Don't forget to balance the equation first. Hope this helps :)

Kevin Antony 2B
Posts: 65
Joined: Sat Sep 07, 2019 12:16 am

Re: Mass Yield of Product

Postby Kevin Antony 2B » Sat Sep 28, 2019 10:11 pm

You forgot to balance the equation. When you do, the equation shows that 1 mole of C6H9Cl3 will yield 3 moles of AgCl. Thus, the .004 moles C6H9Cl3 should yield .012 moles AgCl. Just remember to convert to grams after!

jisulee1C
Posts: 94
Joined: Thu Jul 25, 2019 12:17 am

Re: Mass Yield of Product

Postby jisulee1C » Sat Sep 28, 2019 10:32 pm

After balancing the equation you can use the molar ratio where 1 mol of C6H9Cl3 is equal to 3 moles of AgCl and using this ratio and dimensional analysis will yield you 0.012 moles AgCl which you can then multiply by the molar mass of AgCl to get the final mass of AgCl produced.

Abigail Sanders 1E
Posts: 91
Joined: Wed Sep 11, 2019 12:16 am

Re: Mass Yield of Product

Postby Abigail Sanders 1E » Mon Sep 30, 2019 2:18 pm

Like the previous responses said, your answer was almost right but you forgot to balance the equation beforehand. The balanced equation should be C6H9Cl3 + 3AgNo3 --> 3AgCl + C6H9(NO3)3. Then, once you find that the limiting reactant you must multiple the moles by 3 and then convert to grams. This is because the ratio of C6H9Cl3 to AgCl is 1:3.


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