Question

Is produced by the reaction of 0.4235 mol of CuCl₂ according to the following equation: 2CuCl₂+4KI⟶2CuI+4KCl+I₂.

How many moles of iodine (I₂) are produced in this reaction?
a) 0.2117 mol
b) 0.4235 mol
c) 0.6352 mol
d) 0.8470 mol

Answer 1

In the given reaction, when 0.4235 mol of CuCl₂ reacts, 0.2117 mol of I₂ is produced.

Step-by-step explanation:

The reaction given is:

2CuCl2 + 4KI → 2CuI + 4KCl + I2

From the balanced equation, we can see that 2 moles of CuCl2 produce 1 mole of I2.

Therefore, if 0.4235 mol of CuCl2 is used in the reaction, then the number of moles of I2 produced is 0.4235 mol × (1 mole I2 / 2 moles CuCl2) = 0.2117 mol. So the correct answer is (a) 0.2117 mol.