In the following reaction, calculate the mass of `Cl_(2)` required to produce 1385 g of `KCiO_(4) (K=39):-`
`Cl_(2)+ 2KOH to KCl +KClO +H_(2)O`
`3KClO to 2 KCl +KClO_(3)`
`4KClO_(3)to 3KClO_(4)+ KCl`

Options

A. 40 g
B. 2840 g
C. 1420 g
D. 710 g

Answer

To calculate the mass of \( Cl_2 \) required to produce 1385 g of \( KClO_4 \), we will follow these steps:

### Step 1: Write the balanced reactions
We have the following reactions:
1. \( Cl_2 + 2 KOH \rightarrow KCl + KClO + H_2O \)
2. \( 3 KClO \rightarrow 2 KCl + KClO_3 \)
3. \( 4 KClO_3 \rightarrow 3 KClO_4 + KCl \)

### Step 2: Combine the reactions
To find a single overall reaction, we need to multiply the reactions appropriately and add them together:
- Multiply the first reaction by 12.
- Multiply the second reaction by 4.
- Keep the third reaction as it is.

This gives us:
- From the first reaction: \( 12 Cl_2 + 24 KOH \rightarrow 12 KCl + 12 KClO + 12 H_2O \)
- From the second reaction: \( 4 KClO \rightarrow 2 KCl + KClO_3 \) (multiplied by 4)
- From the third reaction: \( 4 KClO_3 \rightarrow 3 KClO_4 + KCl \)

### Step 3: Add the reactions
When we add these reactions, the \( KClO \) and \( KClO_3 \) will cancel out:
\[ 12 Cl_2 + 24 KOH \rightarrow 3 KClO_4 + 21 KCl + 12 H_2O \]

### Step 4: Calculate the molar masses
- Molar mass of \( KClO_4 \):
- \( K = 39 \, g/mol \)
- \( Cl = 35.5 \, g/mol \)
- \( O = 16 \, g/mol \times 4 = 64 \, g/mol \)

Total: \( 39 + 35.5 + 64 = 138.5 \, g/mol \)

- Molar mass of \( Cl_2 \):
- \( Cl = 35.5 \, g/mol \times 2 = 71 \, g/mol \)

### Step 5: Set up the proportion
From the balanced equation, we see that:
- 12 moles of \( Cl_2 \) produce 3 moles of \( KClO_4 \).

We need to find how many grams of \( Cl_2 \) are required to produce 1385 g of \( KClO_4 \).

### Step 6: Calculate the moles of \( KClO_4 \)
\[ \text{Moles of } KClO_4 = \frac{1385 \, g}{138.5 \, g/mol} = 10 \, moles \]

### Step 7: Calculate the moles of \( Cl_2 \) required
Using the stoichiometry from the balanced equation:
\[ \frac{12 \, \text{moles } Cl_2}{3 \, \text{moles } KClO_4} = \frac{x \, \text{moles } Cl_2}{10 \, \text{moles } KClO_4} \]
Cross-multiplying gives:
\[ x = \frac{12 \times 10}{3} = 40 \, \text{moles of } Cl_2 \]

### Step 8: Calculate the mass of \( Cl_2 \)
\[ \text{Mass of } Cl_2 = 40 \, \text{moles} \times 71 \, g/mol = 2840 \, g \]

### Final Answer
The mass of \( Cl_2 \) required to produce 1385 g of \( KClO_4 \) is **2840 g**.

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