Calculate the limiting reagent : CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl
To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
First, let's write out the balanced equation:
CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl
The stoichiometric ratio between CaCl2 and AgCl is 1:2. This means that 1 mole of CaCl2 reacts with 2 moles of AgCl.
To find the limiting reagent, we need to calculate the moles of each reactant.
For CaCl2:
Given mass of CaCl2 = 5.00 g
Molar mass of CaCl2 = (1 * 40.08 g/mol) + (2 * 35.45 g/mol) = 110.98 g/mol
Moles of CaCl2 = 5.00 g / 110.98 g/mol ≈ 0.045 mol
For AgNO3:
Given mass of AgNO3 = 10.00 g
Molar mass of AgNO3 = (2 * 14.01 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol) = 169.87 g/mol
Moles of AgNO3 = 10.00 g / 169.87 g/mol ≈ 0.059 mol
Now, let's compare the ratio of moles of CaCl2 to AgCl.
The ratio of moles of CaCl2 to AgCl = 0.045 mol : (0.059 mol * (2 mol AgCl / 1 mol CaCl2)) ≈ 0.045 mol : 0.118 mol ≈ 0.382 : 1
Therefore, the limiting reagent is CaCl2.