If you mix 39.19 mL of 2.405 M AgNO3 with 48.91 mL of 0.924 M CaCl2, how many moles of AgCl can you make?

Question

AgNO₃ + CaCl₂ --> AgCl + Ca(NO₃)₂

Answer 1

2AgNO₃ + CaCl₂ --> 2AgCl + Ca(NO₃)₂ .. balanced equation

Since we are given the amounts of BOTH reactants, we must first find which one is limiting. One way to do this to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For AgNO₃: 39.19 ml x 1 L / 1000 ml x 2.405 mol / L = 0.09425 mols AgNO₃(÷2->0.0471)

For CaCl₂: 48.91 ml x 1 L / 1000 ml x 0.924 mol / L = 0.04519 mols CaCl₂(÷1->0.0452)

Since 0.0452 is less than 0.0471, CaCl₂ is limiting and the moles of CaCl₂(0.04519 mols) will determine how much product (AgCl) can be formed.

Using the stoichiometry of the balanced equation along with dimensional analysis, we can now determine the moles of AgCl formed;

0.04591 mols CaCl₂ x 2 mols AgCl / mol CaCl₂ = 0.0918 mols AgCl formed (to 3 sig. figs.)