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2. For the reaction [tex]Cl_2(g) + 2KBr(s) \rightarrow 2KCl(s) + Br_2(g)[/tex], how many grams of potassium chloride can be produced from [tex]300 \text{ g}[/tex] of potassium bromide?

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GinnyAnswer
To solve this problem, let's follow a detailed, step-by-step approach:

### Step 1: Understand the chemical reaction and stoichiometry

The balanced chemical reaction is:
[tex]\[ \text{Cl}_2(g) + 2 \text{KBr}(s) \rightarrow 2 \text{KCl}(s) + \text{Br}_2(g) \][/tex]

From the balanced equation, we see that:
- 2 moles of KBr produce 2 moles of KCl.

### Step 2: Determine molar masses

Next, we need to know the molar masses of the reactants and the products involved:
- Molar mass of potassium bromide (KBr) = 119.002 g/mol.
- Molar mass of potassium chloride (KCl) = 74.5513 g/mol.

### Step 3: Calculate the moles of potassium bromide (KBr)

We are given 300 grams of potassium bromide (KBr). To find the moles of KBr, we use the formula:
[tex]\[ \text{Moles of KBr} = \frac{\text{Mass of KBr}}{\text{Molar mass of KBr}} \][/tex]

Substitute the given values:
[tex]\[ \text{Moles of KBr} = \frac{300 \text{ g}}{119.002 \text{ g/mol}} \approx 2.521 \text{ moles} \][/tex]

### Step 4: Use stoichiometry to find the moles of potassium chloride (KCl)

From the balanced chemical equation, we see that the reaction is a 1:1 mole ratio between KBr and KCl:
- 1 mole of KBr produces 1 mole of KCl.

Therefore, the moles of KCl produced will be the same as the moles of KBr:
[tex]\[ \text{Moles of KCl} = 2.521 \text{ moles} \][/tex]

### Step 5: Calculate the mass of potassium chloride (KCl)

Finally, to find the mass of potassium chloride produced, we use the formula:
[tex]\[ \text{Mass of KCl} = \text{Moles of KCl} \times \text{Molar mass of KCl} \][/tex]

Substitute the values:
[tex]\[ \text{Mass of KCl} = 2.521 \text{ moles} \times 74.5513 \text{ g/mol} \approx 187.941 \text{ g} \][/tex]

### Conclusion
The mass of potassium chloride (KCl) that can be produced from 300 grams of potassium bromide (KBr) is approximately 187.941 grams.
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