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.
