Which equation shows how to calculate how many grams (g) of $ \text{KOH} $ would be needed to fully react with $ 4 \, \text{mol} \, \text{Mg}(\text{OH})_2 $? The balanced reaction is:

[tex]\[ \text{MgCl}_2 + 2 \, \text{KOH} \rightarrow \text{Mg}(\text{OH})_2 + 2 \, \text{KCl} \][/tex]

A. $\frac{4 \, \text{mol} \, \text{Mg}(\text{OH})_2}{1} \times \frac{2 \, \text{mol} \, \text{KOH}}{1 \, \text{mol} \, \text{Mg}(\text{OH})_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}$

B. $\frac{4 \, \text{mol} \, \text{Mg}(\text{OH})_2}{1} \times \frac{1 \, \text{mol} \, \text{KOH}}{2 \, \text{mol} \, \text{Mg}(\text{OH})_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}$

C. $\frac{1 \, \text{mol} \, \text{MgCl}_2}{1} \times \frac{2 \, \text{mol} \, \text{KOH}}{1 \, \text{mol} \, \text{MgCl}_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}$

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

To determine how many grams of $ KOH $ are needed to fully react with $ 4 \, \text{mol} \, Mg(OH)_2 $ according to the balanced reaction:

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

we need to go through the stoichiometric calculations step-by-step.

First, we need to identify the molar relationship given in the balanced equation:
[tex]\[ 2 \, \text{mol} \, KOH \rightarrow 1 \, \text{mol} \, Mg(OH)_2 \][/tex]

We are given $ 4 \, \text{mol} \, Mg(OH)_2 $. Using stoichiometry:
[tex]\[ 4 \, \text{mol} \, Mg(OH)_2 \times \frac{2 \, \text{mol} \, KOH}{1 \, \text{mol} \, Mg(OH)_2} \][/tex]

Next, we calculate the amount of $ KOH $ needed in moles:
[tex]\[ 4 \times 2 = 8 \, \text{mol} \, KOH \][/tex]

Now, we convert the moles of $ KOH $ to grams. The molar mass of $ KOH $ is given as $ 56.10 \, \text{g/mol} $. Using this, we perform the conversion:
[tex]\[ 8 \, \text{mol} \, KOH \times 56.10 \, \text{g/mol} \, KOH \][/tex]

This calculation gives:
[tex]\[ 8 \times 56.10 = 448.8 \, \text{grams} \, KOH \][/tex]

Therefore, the total grams of $ KOH $ needed to fully react with $ 4 \, \text{mol} \, Mg(OH)_2 $ is $ 448.8 \, \text{grams} $.

Upon analyzing the provided options, we find that:

- Option A is the correct equation:
[tex]\[ \frac{4 \, \text{mol} \, Mg(OH)_2}{1} \times \frac{2 \, \text{mol} \, KOH}{1 \, \text{mol} \, Mg(OH)_2} \times \frac{56.10 \, \text{g} \, KOH}{1 \, \text{mol} \, KOH} \][/tex]
This matches our detailed, step-by-step calculation. Thus, [tex]$ 4 \, \text{mol} \, Mg(OH)_2 $[/tex] needs [tex]$ 448.8 \, \text{g} \, KOH $[/tex] to fully react.