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]
