Alright, let's go through the balanced chemical equation and solve the problem step by step:
The given balanced reaction is:
[tex]\[ \text{MgCl}_2 + 2 \text{KOH} \rightarrow \text{Mg(OH)}_2 + 2 \text{KCl} \][/tex]
From the balanced equation, we see that 2 moles of KOH produce 2 moles of KCl. This gives a 1:1 molar ratio between KOH and KCl.
Now, we need to find out how many grams of KCl are produced from 4 moles of KOH.
### Step 1: Convert moles of KOH to moles of KCl
Since the molar ratio of KOH to KCl is 1:1, 4 moles of KOH would produce 4 moles of KCl.
### Step 2: Convert moles of KCl to grams
The molar mass of KCl is given as 74.55 g/mol. To find the grams of KCl produced from the moles of KCl, we use the formula:
[tex]\[ \text{grams} = \text{moles} \times \text{molar mass} \][/tex]
Substitute the values into the equation:
[tex]\[ \text{grams of KCl} = 4 \text{ moles of KCl} \times 74.55 \text{ g/mol} \][/tex]
### Step 3: Analyze the given options
We need to see which option correctly captures the steps we followed above.
Option A:
[tex]\[ \frac{4 \text{ mol KOH}}{1} \times \frac{1 \text{ mol KCl}}{2 \text{ mol KOH}} \times \frac{74.55 \text{ g KCl}}{1 \text{ mol KCl}} \][/tex]
This option uses an incorrect molar ratio, 1 mol KCl per 2 mol KOH, which isn't right since the actual ratio is 1:1.
Option B:
[tex]\[ \frac{4 \text{ mol KOH}}{1} \times \frac{2 \text{ mol KCl}}{2 \text{ mol KOH}} \times \frac{74.55 \text{ g KCl}}{1 \text{ mol KCl}} \][/tex]
This correctly simplifies the molar ratio of KOH to KCl as 1:1 and uses the correct molar mass of KCl.
Option C:
[tex]\[ \frac{4 \text{ mol KOH}}{1} \times \frac{2 \text{ mol KCl}}{1 \text{ mol KOH}} \times \frac{74.55 \text{ g KCl}}{1 \text{ mol KCl}} \][/tex]
This uses an incorrect ratio that implies 2 moles of KCl are produced from 1 mole of KOH.
Option D:
[tex]\[ \frac{4 \text{ mol KOH}}{1} \times \frac{1 \text{ mol KOH}}{1 \text{ mol KCl}} \times \frac{74.55 \text{ g KCl}}{1 \text{ mol KCl}} \][/tex]
This incorrectly inverts the molar ratio, suggesting a 1:1 ratio of KOH to KOH, which is not what we need for the conversion.
Thus, Option B:
[tex]\[ \frac{4 \text{ mol KOH}}{1} \times \frac{2 \text{ mol KCl}}{2 \text{ mol KOH}} \times \frac{74.55 \text{ g KCl}}{1 \text{ mol KCl}} \][/tex]
is correct.
This matches our steps and provides us with the correct method to calculate the grams of KCl from 4 moles of KOH.
