What is the mass of AgCl produced?

[Chem_Mod]

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 don't know what to do after I convert to grams to mols. Can I get a walkthrough of the rest of the problem?

[Chem_Mod]

First make sure chemical equation is balanced: C6H9Cl3 + 3AgNO3 ---> 3AgCl + C6H9(NO3)3. Then convert the 0.750 g of C6H9Cl3 to moles. Then convert the 1.000 kg of AgNO3 to moles. Then, based on the 1:3 molar ratio in the balanced chemical equation, determine which reactant is the limiting reactant and use its moles to determine how many moles of AgCl can be produced. Then convert the moles of AgCl produced to mass of AgCl produced.

[Kevin Hernandez 3A]

After calculating the moles, how do you determine which is the limiting reactant. I know you have to compare the ratios however, I am not sure on what I'm looking for.

[Pooja Nair 1C]

After you calculate the moles of each, compare the moles of C6H9Cl3 and moles of AgNO3. Since the balanced equation has a 1:3 ratio, the number of moles of AgNO3 should be at least three times more than the number of moles of C6H9Cl3. If there are less than 3x, then AgNO3 is the limiting reactant because there is not enough required. If there are more than 3x, then C6H9Cl3 is the limiting reactant, because after the reaction is completed there is leftover AgNO3. After determining the limiting reactant, you multiply the number of moles at AgCl by the moles of the limiting reactant to find the moles of product.

[SammiOrsini_1B]

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?
This question is in the module on Limiting Reactant Calculations. I have followed the steps provided however my answer does not match up with any of the answer choices. I converted the 0.750 g of C6H9Cl3 to moles, and then converted the 1.000 kg of AgNO3 to moles. Then I used the 1:3 molar ratio in the balanced chemical equation to determine which reactant is the limiting reactant and then used its moles to determine how many moles of AgCl can be produced. After i convert the moles of AgCl to grams my answer does not correlate with the given answers. Does anyone know what I'm doing wrong?

[SammiOrsini_1B]

After guessing a checking i found that the answer is 1.72 g AgCl. I found that the limiting reactant is C6H9Cl3 and that the moles of this compound is .004 moles. To get the answer 1.72, i saw that you have to triple .004 moles and then multiple that by the molar mass which is 143.32 g/mol. I am unclear why you need to triple .004 moles. Why cant we use the .004 moles, then multiple this by the molar mass?

[Jocelyn Fermin1J]

I think you have to triple the .004 because of the 3 in front of AgNO3.

[Yadira Flores 1G]

I am stuck on this problem too. I found the mol of C6H9Cl3 to be 0.004 and then continued by multiplying that my 3molAgNO3/1molC6H9Cl3 and got 0.012molAgNO3 so then I multiplied that by the molar mass 169.88g of AgNO3 but my answer isn't correct. Can anyone help?

[Caitlyn Ponce 1L]

In order to get the mass of AgCl, you have to multiply the produced moles of AgCl, which is 0.012 moles, with the molar mass of AgCl, which is 143.32 g/mol. In doing so, the answer will be 1.72 g AgCl. I think you got mixed up and used AgNO3 instead, but the question is asking for the mass of AgCl. We multiply 0.004 moles with 3 because the ratio of C6H9Cl3 to AgCl is 1:3 (the equation must be balanced first in order to see this ratio).

[John Kim Lec3Dis3L]

After guessing a checking i found that the answer is 1.72 g AgCl. I found that the limiting reactant is C6H9Cl3 and that the moles of this compound is .004 moles. To get the answer 1.72, i saw that you have to triple .004 moles and then multiple that by the molar mass which is 143.32 g/mol. I am unclear why you need to triple .004 moles. Why cant we use the .004 moles, then multiple this by the molar mass?

Actually, the answer to your question shows up earlier in the discussion, post 2 to be specific. Basically, you have to balance the equation before you can calculate the mass of AgCl. In the case of this chemical reaction, you can see that the stoichiometric coefficient for AgNO3 is three; to balance the equation, there needs to be three Ag particles present on the product side as well.

[RoopshaChatterjee 1G]

I'm having some difficulty with a similar question: Consider the unbalanced reaction: PCl3+ H2O --->HCl + H3PO3 What mass of HCl is produced by the reaction of 23.6g of PCl3? Do you find the mass of HCl through a similar method?