Posted by Michelle on .
158.5 mL of a AgNO3 solution at 5.0M was combined with a 3.5M CaCl2 solution. 110.5g of AgCl was recovered. Given that the CaCl2 is the limiting reagent, how many milliliters of the 3.5M CaCl2 solution was used? 2 sig figs
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
mole AgCl = 110.5/molar mass = ??
moles CaCl2 used = moles AgCl x (1 mole CaCl2/2 moles AgCl) = ??
M CaCl2 = moles CaCl2/L CaCl2.
Solve for L CaCl2 and convert to mL.
All of this depends, of course, upon the reaction being complete and recovery being 100% of the theoretical yield.