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

B. [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]

C. [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]

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 :

GinnyAnswer
To find out how many grams of [tex]\( Mg(OH)_2 \)[/tex] would be produced from 4 mol of [tex]\( KOH \)[/tex], we start with the balanced chemical equation:

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

From the balanced equation, we see that 2 moles of [tex]\( KOH \)[/tex] produce 1 mole of [tex]\( Mg(OH)_2 \)[/tex]. To calculate the grams of [tex]\( Mg(OH)_2 \)[/tex] produced:

1. Determine the moles of [tex]\( Mg(OH)_2 \)[/tex] produced:

Given 4 moles of [tex]\( KOH \)[/tex], and the stoichiometric ratio (2 moles [tex]\( KOH \)[/tex] to 1 mole [tex]\( Mg(OH)_2 \)[/tex]):
[tex]\[ 4 \text{ mol } KOH \times \frac{1 \text{ mol } Mg(OH)_2}{2 \text{ mol } KOH} = 2 \text{ mol } Mg(OH)_2 \][/tex]

2. Calculate the grams of [tex]\( Mg(OH)_2 \)[/tex] produced:

The molar mass of [tex]\( Mg(OH)_2 \)[/tex] is 58.31 g/mol. Using the moles of [tex]\( Mg(OH)_2 \)[/tex] we calculated:
[tex]\[ 2 \text{ mol } Mg(OH)_2 \times 58.31 \text{ g/mol } = 116.62 \text{ g } Mg(OH)_2 \][/tex]

Thus, option C correctly matches the process:
[tex]\[ \frac{4 mol KOH }{1} \times \frac{1 mol Mg(OH)_2}{2 mol KOH } \times \frac{58.31 g Mg(OH )_2}{1 mol Mg(OH)_2} \][/tex]

Therefore, the correct equation is:
C. [tex]\(\frac{4 mol KOH}{1} \times \frac{1 mol Mg(OH )_2}{2 mol KOH} \times \frac{58.31 g Mg(OH )_2}{1 mol Mg(OH )_2}\)[/tex]