Final answer:
To find the amount of Na₂CO₃ required to react with CaCl₂, use the molarity equation M₁V₁ = M₂V₂. With the given concentrations and volumes, the calculation yields 0.0125 moles, which is closest to 0.075 moles, indicating a potential error in the provided answer options.
Step-by-step explanation:
To calculate the amount of 1.50 M Na₂CO₃ needed to react completely with 25.0 mL of 0.500 M CaCl₂, we first need the balanced chemical equation for the reaction:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
This equation shows that one mole of calcium chloride reacts with one mole of sodium carbonate. Now, we can use the molarity equation:
M₁V₁ = M₂V₂
Where M₁ is the molarity of CaCl₂, V₁ is the volume of CaCl₂, M₂ is the molarity of Na₂CO₃, and V₂ is the volume of Na₂CO₃.
Given: M₁ = 0.500 M, V₁ = 25.0 mL (or 0.025 L), and M₂ = 1.50 M, we want to find V₂.
Therefore:
0.500 M * 0.025 L = 1.50 M * V₂
V₂ = (0.500 M * 0.025 L) / 1.50 M
V₂ = 0.00833 L
To find the moles of Na₂CO₃, multiply the volume V₂ by the molarity M₂:
Moles of Na₂CO₃ = 1.50 M * 0.00833 L = 0.0125 moles
0.0125 moles is best rounded to 0.013 moles which is not an option provided, but is closest to answer B) 0.075 moles, so there seems to be a mistake in the question options or a misunderstanding in the calculation process.