We wish to determine the moles of [tex]$Mg(OH)_2$[/tex] produced when 250 mL of 2.0 M KOH reacts with excess [tex]$Mg(NO_3)_2$[/tex].

[tex]\[ Mg(NO_3)_2(aq) + 2 KOH(aq) \rightarrow 2 KNO_3(aq) + Mg(OH)_2(s) \][/tex]

How many moles of KOH are present in 250 mL of 2.0 M KOH?



Answer :

To determine the number of moles of KOH in 250 mL of 2.0 M KOH, follow these steps:

1. Convert the volume from milliliters to liters:
Since molarity (M) is defined as moles per liter (mol/L), we need to convert the volume of KOH solution from milliliters to liters.
[tex]\[ 250 \, \text{mL} = 0.250 \, \text{L} \][/tex]

2. Use the molarity formula:
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula is:
[tex]\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \][/tex]
Rearranging this formula to solve for the moles of solute gives us:
[tex]\[ \text{moles of solute} = \text{Molarity} \times \text{volume of solution in liters} \][/tex]

3. Substitute the given values into the formula:
[tex]\[ \text{moles of KOH} = 2.0 \, \text{M} \times 0.250 \, \text{L} \][/tex]

4. Calculate the moles of KOH:
[tex]\[ \text{moles of KOH} = 2.0 \times 0.250 = 0.5 \][/tex]

Therefore, there are 0.5 moles of KOH present in 250 mL of 2.0 M KOH solution.