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]
[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]
