In the reaction of 423 g of CuCl2, approximately 1.575 moles of I2 are produced. The mass of I2 formed is approximately 399.23 g.
To solve this problem, we'll follow these steps:
1. Calculate the moles of CuCl2 using its molar mass.
2. Use the balanced chemical equation to find the moles of I2 produced in the reaction.
3. Calculate the molar mass of I2 to find the mass.
Let's proceed with the calculations:
**Step 1: Calculate moles of CuCl2**
\[ \text{Molar mass of } CuCl_2 = (63.55 + 2 \times 35.45) \, \text{g/mol} = 134.45 \, \text{g/mol} \]
\[ \text{Moles of } CuCl_2 = \frac{423 \, \text{g}}{134.45 \, \text{g/mol}} \]
\[ \text{Moles of } CuCl_2 \approx 3.15 \, \text{mol} \]
**Step 2: Use the balanced chemical equation**
\[ 2 \, \text{CuCl}_2 + 4 \, \text{KI} \rightarrow 2 \, \text{CuI} + 4 \, \text{KCl} + \text{I}_2 \]
From the balanced equation, 2 moles of CuCl2 produce 1 mole of I2.
\[ \text{Moles of } I_2 = \frac{3.15 \, \text{mol}}{2} \]
\[ \text{Moles of } I_2 \approx 1.575 \, \text{mol} \]
**Step 3: Calculate mass of I2**
\[ \text{Molar mass of } I_2 = 2 \times 126.90 \, \text{g/mol} = 253.80 \, \text{g/mol} \]
\[ \text{Mass of } I_2 = 1.575 \, \text{mol} \times 253.80 \, \text{g/mol} \]
\[ \text{Mass of } I_2 \approx 399.23 \, \text{g} \]
Answer:
1. Approximately 1.575 moles of I2 are produced.
2. Approximately 399.23 grams of I2 are formed.