Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Which equation shows how to calculate how many grams (g) of KCl would be produced from 4 mol KOH? The balanced reaction is:
[tex]\[
MgCl_2 + 2 KOH \rightarrow Mg(OH)_2 + 2 KCl
\][/tex]

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]

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]

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]

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]


Sagot :

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.