Question

which equation shows how to calculate how many grams (g) of mg(oh)₂ would be produced from 4 mol koh? the balanced reaction is: mgcl₂ + 2koh → mg(oh)₂ + 2kcl
a. $\frac{4 mol koh}{1}\times\frac{1 mol mg(oh)_2}{2 mol koh}\times\frac{58.31 g mg(oh)_2}{1 mol mg(oh)_2}$
b. $\frac{4 mol koh}{1}\times\frac{2 mol koh}{1 mol mg(oh)_2}\times\frac{56.10 g koh}{1 mol koh}$
c. $\frac{4 mol koh}{1}\times\frac{2 mol mg(oh)_2}{1 mol koh}\times\frac{58.31 g mg(oh)_2}{1 mol mg(oh)_2}$
d. $\frac{4 mol koh}{1}\times\frac{1 mol mg(oh)_2}{1 mol koh}\times\frac{58.31 g mg(oh)_2}{1 mol mg(oh)_2}$

Explanation:

Step1: Determine mole - ratio

From the balanced equation $MgCl_2 + 2KOH
ightarrow Mg(OH)_2+2KCl$, the mole - ratio of $KOH$ to $Mg(OH)_2$ is $2:1$. So, for every 2 moles of $KOH$, 1 mole of $Mg(OH)_2$ is produced. To convert moles of $KOH$ to moles of $Mg(OH)_2$, we use the conversion factor $\frac{1\ mol\ Mg(OH)_2}{2\ mol\ KOH}$.

Step2: Convert moles of $Mg(OH)_2$ to grams

The molar mass of $Mg(OH)_2$ is approximately $58.31\ g/mol$. To convert moles of $Mg(OH)_2$ to grams, we use the conversion factor $\frac{58.31\ g\ Mg(OH)_2}{1\ mol\ Mg(OH)_2}$.

Answer:

A. $\frac{4\ mol\ KOH}{1}\times\frac{1\ mol\ Mg(OH)_2}{2\ mol\ KOH}\times\frac{58.31\ g\ Mg(OH)_2}{1\ mol\ Mg(OH)_2}$