Question on Post-Module #22

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ariellasarkissian_3H
Posts: 46
Joined: Fri Sep 28, 2018 12:25 am

Question on Post-Module #22

Postby ariellasarkissian_3H » Mon Oct 08, 2018 8:38 pm

After converting grams of C6H9Cl3 and AgNO3 to moles, I got 0.004 moles and 17.659 moles, respectively. After this step, I used the limiting reactant moles (C6H9Cl3) and converted it to grams of product (AgCl). The answer was 0.57328, but this was not close to the real answer. Does anyone know the steps to get the correct answer?

bonnie_schmitz_1F
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Joined: Fri Sep 28, 2018 12:24 am
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Re: Question on Post-Module #22

Postby bonnie_schmitz_1F » Mon Oct 08, 2018 8:51 pm

I got the same answer as you and was also unsure how to get the correct answer here. Something I was wondering about was the units of the reactants: one was 0.750 g of C6H9Cl3 which is grams and the other, 1.000 kg of AgNO3, is in kilograms. I thought there might be a typo and they both should have been in kg, so I tried that but still did not get the correct answer.

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: Question on Post-Module #22

Postby ChathuriGunasekera1D » Mon Oct 08, 2018 8:58 pm

Hi! I think the equation wasn't fully balanced. This is what is given:

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

But there are 3 moles of Ag on the left and only 1 mole on the right. There should be a 3 in front of the AgCl. That means if you multiply your 0.5732 mol by 3, you get 1.72 g and that is one of the answers!


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