Question

If 30.0mL of 0.150M CaCl2 is added to 15.0mL of 0.100M AgNO3, what is the mass in grams of AgCl participate?

Answer 1

The first step is to make a balanced chemical equation.
 2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2

Molecular Weights:
CaCl2 = 110.98 g/mol
AgNO3 =170.01
AgCl: 143.45 g/mol

Volume:
CaCl2: 30.0mL=0.03L
AgNO3: 15.0mL=0.015 L


Solving for the limiting reactant one needs to get the mols CaCl2 and mols AgNO3:

CaCl2: 0.150M(mol/L) * 0.03L = 0.0045 moles
AgNO3: 0.100M*0.015L = 0.0015 moles

Since the stoichiometric ratio of AgNO3 to CaCl2 is 2:1
0.0015 mols AgNO3 *(1 mol CaCl2/ 2 mols AgNO3) = 0.00075 mols CaCl2

Since the answer is lesser than CaCl2 then the limiting reactant is AgNO3.

To get the mass of AgCl one will do a stoichiometric calculation with respect to the limiting reactant, AgNO3.

0.0015 moles AgNO3 * $( \frac{2 mols AgCl}{2 mols AgNO3}) *( \frac{143.45 g AgCl}{mol AgCl}) = 0.215 moles AgNO3$