Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.?
Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.
a) How many moles of AgCl would precipitate?
b) What are the concentrations in M of Ag+, Cl-, NO3- and Na+ in the reaction mixture after the reaction occurs?
- ?Lv 72 months agoFavourite answer
Moles of Cl- = 0.490 moles / L *0.0267 L = 0.01308 moles = moles Na+
moles of Ag+ = 0.225 mole/L * 0.0267 L = 0.00601 moles = moles NO3-
Ag++Cl- = AgCl
according to the equation we have a 1:1 ratio of reactants
we also see that Ag+ is the limiting reactant
so we will get 0.00601 moles of AgCl
the Ag+ will all precipitate so very little left in solution
moles of Cl- remaining = 0.01308 moles - 0.00601 moles = 0.00708 moles
these are in 2* 0.0267 L = 0.0534 L
[Cl-] = 0.00708 moles / 0.0534 L = 0.132 M
[NO3] = 0.00601 moles / 0.0534 L = 0.112 M
[Cl-] = 0.01308 moles / 0.0534 L = 0.245 M