Answered

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

[tex]\[ MgCl_2 + 2 KOH \rightarrow Mg(OH)_2 + 2 KCl \][/tex]

A. [tex]\(\frac{4 \text{ mol } KOH}{1} \times \frac{1 \text{ mol } Mg(OH)_2}{1 \text{ mol } KOH} \times \frac{58.31 \text{ g } Mg(OH)_2}{1 \text{ mol } Mg(OH)_2}\)[/tex]

B. [tex]\(\frac{4 \text{ mol } KOH}{1} \times \frac{1 \text{ mol } Mg(OH)_2}{2 \text{ mol } KOH} \times \frac{58.31 \text{ g } Mg(OH)_2}{1 \text{ mol } Mg(OH)_2}\)[/tex]

C. [tex]\(\frac{4 \text{ mol } KOH}{1} \times \frac{2 \text{ mol } KOH}{1 \text{ mol } Mg(OH)_2} \times \frac{56.10 \text{ g } KOH}{1 \text{ mol } KOH}\)[/tex]

D. [tex]\(\frac{4 \text{ mol } KOH}{1} \times \frac{2 \text{ mol } Mg(OH)_2}{1 \text{ mol } KOH} \times \frac{58.31 \text{ g } Mg(OH)_2}{1 \text{ mol } Mg(OH)_2}\)[/tex]



Answer :

To determine the correct equation to calculate how many grams of [tex]\( Mg(OH)_2 \)[/tex] would be produced from 4 mol KOH, let's carefully consider each option in the context of the balanced chemical equation:

[tex]\[ MgCl_2 + 2 KOH \rightarrow Mg(OH)_2 + 2 KCl \][/tex]

We know from the balanced equation that:

- 2 moles of KOH react with 1 mole of [tex]\( Mg(OH)_2 \)[/tex].

### Option Analysis:

Option A:
[tex]\[ \frac{4 \, \text{mol KOH}}{1} \times \frac{1 \, \text{mol} \, Mg(OH)_2}{1 \, \text{mol KOH}} \times \frac{58.31 \, \text{g} \, Mg(OH)_2}{1 \, \text{mol} \, Mg(OH)_2} \][/tex]

- This option uses a 1:1 molar ratio between [tex]\( KOH \)[/tex] and [tex]\( Mg(OH)_2 \)[/tex], which is incorrect because the correct ratio is 2:1.

Option B:
[tex]\[ \frac{4 \, \text{mol KOH}}{1} \times \frac{1 \, \text{mol} \, Mg(OH)_2}{2 \, \text{mol KOH}} \times \frac{58.31 \, \text{g} \, Mg(OH)_2}{1 \, \text{mol} \, Mg(OH)_2} \][/tex]

- This option correctly uses the 2:1 mole ratio (2 mol KOH to 1 mol [tex]\( Mg(OH)_2 \)[/tex]) and proceeds to convert moles of [tex]\( Mg(OH)_2 \)[/tex] to grams using the molar mass of [tex]\( Mg(OH)_2 \)[/tex].

Option C:
[tex]\[ \frac{4 \, \text{mol KOH}}{1} \times \frac{2 \, \text{mol KOH}}{1 \, \text{mol} \, Mg(OH)_2} \times \frac{56.10 \, \text{g} \, KOH}{1 \, \text{mol KOH}} \][/tex]

- This option inverts the molar ratio (incorrectly using 2 mol KOH:1 mol [tex]\( Mg(OH)_2 )), and then erroneously converts moles of KOH instead of \( Mg(OH)_2 \)[/tex] to grams.

Option D:
[tex]\[ \frac{4 \, \text{mol KOH}}{1} \times \frac{2 \, \text{mol} \, Mg(OH)_2}{1 \, \text{mol KOH}} \times \frac{58.31 \, \text{g} \, Mg(OH)_2}{1 \, \text{mol} \, Mg(OH)_2} \][/tex]

- This option also uses an incorrect molar ratio (2 mol [tex]\( Mg(OH)_2 \)[/tex]:1 mol KOH) and thus is incorrect.

### Conclusion:
Option B is the correct equation to calculate how many grams of [tex]\( Mg(OH)_2 \)[/tex] would be produced from 4 mol KOH. It correctly applies the stoichiometric conversion from moles of KOH to moles of [tex]\( Mg(OH)_2 \)[/tex] and then converts to grams.

Thus, the correct equation is:

[tex]\[ \frac{4 \, \text{mol KOH}}{1} \times \frac{1 \, \text{mol Mg(OH)}_2}{2 \, \text{mol KOH}} \times \frac{58.31 \, \text{g Mg(OH)}_2}{1 \, \text{mol Mg(OH)}_2} \][/tex]

So the answer is:
[tex]\[ B \][/tex]