Question

If 30.0 ml of 0.150 m cacl2 is added to 15.0 ml of 0.100 m agno3, what is the mass of the agcl precipitate?

Answer 1

The balanced chemical equation is

 2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2

Molecular Weight of CaCl2 = 110.98 g/mol

AgNO3 =170.01

AgCl: 143.45 g/mol

Volume of 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 *