The first step is to find the number of moles of calcium iodide used in the reaction:
147 g CaI₂ x (1 mol CaI₂/ 214 g CaI₂) = 0.687 mol CaI₂
From the balanced chemical equation, we can see that 1 mole of CaI₂ produces 1 mole of CaCl₂. Therefore, we will have:
0.687 mol CaCl₂ produced
Finally, we can find the mass of CaCl₂ produced by multiplying the number of moles by the molar mass of CaCl₂:
0.687 mol CaCl₂ x 110 g/mol CaCl₂ = 75.6 g CaCl₂
Therefore, the answer is option (b) 55 g.
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?
a) 35 g
b) 55 g
c) 147 g
d) 127 g
1 answer