If you mix 67 mL of 1.375 M AgNO3 with 18.867 mL of 1.238 M CaCl2, how many mol of Ca(NO3)2 can you make?

Question

AgNO₃ + CaCl₂ --> AgCl + Ca(NO₃)₂

Answer 1

This is a stoichiometry and a limiting reactant problem. We want to convert both of our starting quantities into moles of the product, and see which is lesser. Since this is the reactant that would run out first, we should base our further calculations off of it.

67mL (1 L / 1000 mL) (1.375 mol / L) (1 mol Ca(NO₃)₂ / 1 mol AgNO₃) = .092 mol Ca(NO₃)₂

18.867 mL (1 L / 1000 mL) (1.238 mol / L) (1 mol Ca(NO₃)₂ / 1 mol CaCl₂) = .023 mol Ca(NO₃)₂

As we can see, we get that the CaCl₂ will produce less of the product and is therefore the limiting reactant for this problem. We then say that this is the amount of Ca(NO₃)₂ that will be produced because after this amount is produced, there will be no more CaCl₂ to react and therefore no more reaction will happen. Therefore, our final answer should be .023 moles of Ca(NO₃)₂