Use the chemical equation to answer the question.
CaI₂ (s) + Cl₂(g) → CaCl₂(s) + I₂(s)
The molar mass of calcium (Ca) is 40 g/mol, the molar mass of iodine (I) is 127 g/mol, and the molar mass of chlorine (Cl) is 35 g/mol. In a reaction that uses 147 grams of calcium iodide (CaI₂), how many grams of calcium chloride (CaCl₂) will be produced?
Responses:
a) 35 g
b) 55 g
c) 147 g
d) 127 g
First, we need to find the molar mass of calcium iodide (CaI₂) and calcium chloride (CaCl₂).
For CaI₂:
Molar mass of CaI₂ = (1 × 40 g/mol Ca) + (2 × 127 g/mol I) = 40 g/mol + 254 g/mol = 294 g/mol CaI₂
For CaCl₂:
Molar mass of CaCl₂ = (1 × 40 g/mol Ca) + (2 × 35 g/mol Cl) = 40 g/mol + 70 g/mol = 110 g/mol CaCl₂
Now, we need to find the number of moles of CaI₂ used:
Number of moles of CaI₂ = mass / molar mass = 147 g / 294 g/mol = 0.5 mol CaI₂
According to the balanced chemical equation, 1 mole of CaI₂ produces 1 mole of CaCl₂. So, 0.5 mol of CaI₂ will produce 0.5 mol of CaCl₂.
Finally, we need to find the mass of CaCl₂ produced:
Mass of CaCl₂ = number of moles × molar mass = 0.5 mol × 110 g/mol = 55 g
So, the correct answer is:
b) 55 g