Calculate the limiting reagent : CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl

Answer 1

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.