Answer:
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%.
(where Q is heat, m is mass, c is specific heat and 
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature