a. The limiting reactant is Ca(OH)₂
b. The theoretical yield of CaCl₂ is approximately 621.488 grams
c. The percentage yield of CaCl₂ is approximately 47.06%
Explanation:
a. The given chemical reaction is presented as follows;
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
Therefore;
One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O
The mass of HCl in an experiment, m₁ = 229.70 g
The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g
The molar mass of HCl, MM₁ = 36.458 g/mol
The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol
The number of moles of HCl, present, n₁ = m₁/MM₁
∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles
The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂
∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles
The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles
According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl
Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant
b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂
The molar mass of CaCl₂ = 110.98 g/mol
The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams
The theoretical yield of CaCl₂ ≈ 621.488 grams
c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;
The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100
∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%
The percentage yield of CaCl₂ ≈ 47.06%.
