Which of the following setups is correct to calculate the mass (in g) of `KCIO_3` produced from the reacting of `0.150` moles of `CI_2` ?

Options

A. `0.150 " " "moles" " "Cl_(2)xx 1 "mole" " "KClO_(3)//3 "moles" Cl_(2)xx122.5 g//1 " mole" " "KClO_(3)`
B. `0.150" " "moles" " "Cl_(2)xx 1 "mole" " " KClO_(3)//3 "moles" Cl_(2)xx 1"mole" " "KClO_(3)//122.5 g`
C. `0.150 " moles" " "Cl_(2)xx 3 "moles" " "Cl_(2)// 1 " mole " KCLO_(3)xx122.5 g//1 "mole" " "KClO_(3)`
D. `0.150 " moles" " "Cl_(2)xx 3 "moles" " "Cl_(2)// 1 " mole " KCLO_(3)xx1 "mole" " "KClO_(3)// 122.5 g`

Answer

To calculate the mass of KClO3 produced from the reaction of 0.150 moles of Cl2, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{Cl}_2 + 6 \text{KOH} \rightarrow 5 \text{KCl} + \text{KClO}_3 + 3 \text{H}_2\text{O} \]

### Step 2: Determine the mole ratio
From the balanced equation, we can see that:
- 3 moles of Cl2 produce 1 mole of KClO3.

### Step 3: Calculate moles of KClO3 produced
Using the mole ratio, we can find out how many moles of KClO3 are produced from 0.150 moles of Cl2:
\[ \text{Moles of KClO3} = 0.150 \, \text{moles of Cl2} \times \frac{1 \, \text{mole of KClO3}}{3 \, \text{moles of Cl2}} = 0.050 \, \text{moles of KClO3} \]

### Step 4: Calculate the mass of KClO3
Now, we need to convert moles of KClO3 to grams. The molar mass of KClO3 is approximately 122.5 g/mol. Therefore:
\[ \text{Mass of KClO3} = \text{moles of KClO3} \times \text{molar mass of KClO3} \]
\[ \text{Mass of KClO3} = 0.050 \, \text{moles} \times 122.5 \, \text{g/mol} = 6.125 \, \text{g} \]

### Final Answer:
The mass of KClO3 produced from the reaction of 0.150 moles of Cl2 is **6.125 grams**.

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