a. Number of molecules of I2 ≈ \(2.40 \times 10^{23}\) molecules
b. Mass of I2 ≈ 101.41 g
To solve this problem, we'll follow these steps:
**Given:**
- Moles of CuCl2 = 0.3993 mol
- Balanced chemical equation: \(2CuCl_2 + 4KI \rightarrow 2CuI + 4KCl + I_2\)
**a. Number of Molecules of I2 Produced:**
- According to the stoichiometry of the reaction, 1 mole of CuCl2 produces 1 mole of I2.
- Therefore, 0.3993 mol of CuCl2 will produce 0.3993 mol of I2.
- To find the number of molecules, we use Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol).
\[ \text{Number of molecules of I2} = \text{Moles of I2} \times \text{Avogadro's number} \]
\[ \text{Number of molecules of I2} = 0.3993 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \]
**b. Mass of I2 Produced:**
- To find the mass of I2 produced, we use the molar mass of iodine (I2), which is approximately 253.8 g/mol.
\[ \text{Mass of I2} = \text{Moles of I2} \times \text{Molar mass of I2} \]
\[ \text{Mass of I2} = 0.3993 \, \text{mol} \times 253.8 \, \text{g/mol} \]
**Calculations:**
a. \[ \text{Number of molecules of I2} = 0.3993 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \]
b. \[ \text{Mass of I2} = 0.3993 \, \text{mol} \times 253.8 \, \text{g/mol} \]
**Results:**
a. Number of molecules of I2 ≈ \(2.40 \times 10^{23}\) molecules
b. Mass of I2 ≈ 101.41 g