Limiting Reactant Calculations

[Tiffany_Cacy_3D]

21. 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)

This is a question from the Limiting reactant calculations post-module and the other problems were okay for me but for some reason I was having trouble finding this answer. I keep getting extremely big solutions and I am not sure what I am doing wrong. I appreciate the help!

[RachaelKoh3A]

Hi!

Since 1.000kg of AgNO3 is used, C6H9Cl3 is the limiting reactant as AgNO3 is in excess.
Find the number of moles of C6H9Cl3 by taking 0.750 divided by the Mr 187.50 g/mol.
Number of moles of C6H9Cl3 = 0.750/187.50 = 0.00400 mol

The number of moles of AgCl produced is the same as the number of moles of C6H9Cl3 used as per the equation provided.
So number of moles of AgCl = 0.00400 mol

Then to find the mass of AgCl, take the number of moles multiplied by the Mr 143.32 g/mol.
Mass of AgCl produced = 0.00400 x 143.32 = 0.573 g

[Fanny Lee 2K]

First start of by balancing the equation and you should get C6H9Cl3 + 3AgNO3 --> 3AgCl + C6H9(NO3)3. Make sure the given masses of the reactants are converted into grams. In this case, 1.000 kg is equal to 1000 grams. Convert the grams to moles by dividing it by the molar mass. Determine the limiting reactant (in this case is C6H9Cl3) and compare the ratio of that to AgCl in the balanced equation. For every 1 mole of C6H9Cl3 there is 3 AgCl. Take the moles of C6H9Cl3 and multiply it by 3. With the moles of AgCl, convert it into grams since it is asking for the mass by multiplying it by its molar mass. You should get 1.72 grams of AgCl.