Video 3 Post- assessment

[huntercrespo1C]

750 g of CHH9Cl13 is mixed with 1.0 kg of AgNO3 in a flask of water. AgCl precipitates out - what is the mass of AgCl.

[Beverly Shih 1K]

There was one partially confusing piece of information provided on that question that changed my answer the first time around--the molar mass provided for 3AgNO3 (169.88 g/mol) is for one mol of AgNO3, not 3. With that in mind:

Find the moles of C6H9Cl3 by dividing its mass by mass per mole:
0.750g / 187.50 g x mol^-1 = 0.00400 mol C6H9Cl3

Find the moles of Cl by multiplying the moles of C6H9Cl3 by 3 (because there are 3 Cl atoms per C6H9Cl3 molecule):
0.00400 mol x 3 = 0.012 mol Cl

Find the moles of AgNO3 by converting its mass to grams and dividing it by mass per mole:
1.000kg x (1000g / 1 kg) x (1 / 169.88 g x mol^-1) = 5.887 mol AgNO3

Find the moles of Ag, which is the same as moles of AgNO3:
5.887 mol Ag

Determine which reactant is the limiting reactant:
0.0120 mol Cl would produce 0.0120 mol AgCl because you need 1 mol Cl for each mol of AgCl
5.887 mol Ag would produce 5.887 mol / 3 = 1.962 mol AgCl because you need 3 mol Ag for each mol of AgCl
0.0120 < 1.962, so Cl is the limiting reactant which limits the product to 0.0120 mol AgCl

Calculate the mass of AgCl by multiplying moles of AgCl by molar mass:
0.0120 mol x 143.22 g x mol-1 = 1.72 g AgCl

Hope this helps!