## Q.22 on Audio-Visual Focus-Topic [ENDORSED]

Mariah Guerrero 1J
Posts: 32
Joined: Fri Apr 06, 2018 11:03 am

### Q.22 on Audio-Visual Focus-Topic

I am reviewing the post-assessment for the audio-visual focus-topic for limiting reactants, and I am stuck on this 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
Molar Mass: C6H9Cl3 (187.50 g/mol), 3AgNO3 (169.88 g/mol), AgCl (143.32 g/mol)

I calculated the moles of each reactant and found that the limiting reactant is C6H9Cl3, but I'm not sure what to do next. Any suggestions?

Marisol Sanchez - 1E
Posts: 38
Joined: Fri Apr 06, 2018 11:05 am

### Re: Q.22 on Audio-Visual Focus-Topic

Once you have found which reactant is limiting, you would follow by finding how much of the product can be made given the mass of the reactant. You would use the molar ratios of the reactant and the product so make sure you convert the mass of the reactant from grams to mols. Once that part is solved, this will give you the amount in mols of the product and then you can convert it by using its molar mass to get the actual mass.

Andre-1H
Posts: 39
Joined: Fri Apr 06, 2018 11:01 am

### Re: Q.22 on Audio-Visual Focus-Topic  [ENDORSED]

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

You just use the reaction and the stoichiometric coefficients to determine how many moles can be produced. So you said you found C6H9Cl3 was the limiting reactant and by doing that you did the hardest part of the problem, now you use how many moles of that to see how many moles of AgCl you can produce. The equation tells you that for every 1 mole of C6H9Cl3 1 mole of AgCl is produced so then just multiply however many moles of AgCl by its molar mass to find how much in grams would be produced.