How many milliliters of 0.316 M KOH are needed to react completely with 58.0 mL of 0.297 M FeCl2 solution to precipitate Fe(OH)2? The net ionic equation is:
Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
mmols FeCl2 = mL x M = ?
mmols KOH = 2x mmols FeCl2
M KOH = mmols KOH/mL KOH. You know mmols kOH and M KOH, solve for mL KOH.
To find out how many milliliters of 0.316 M KOH solution are needed to react completely with 58.0 mL of 0.297 M FeCl2 solution, we can use the stoichiometry of the reaction and the concept of molarity.
The net ionic equation shows that 1 mole of Fe2+ ions reacts with 2 moles of OH- ions to form 1 mole of Fe(OH)2. Based on this information, we can set up a stoichiometric ratio:
1 mole Fe2+ : 2 moles OH- : 1 mole Fe(OH)2
Next, let's calculate the number of moles of Fe2+ and OH- in the given 58.0 mL of 0.297 M FeCl2 solution:
moles of Fe2+ = volume (L) x molarity
moles of Fe2+ = 0.058 L x 0.297 mol/L
Similarly, we can calculate the moles of OH- ions needed to react with Fe2+ ions:
moles of OH- = 2 x moles of Fe2+
moles of OH- = 2 x (0.058 L x 0.297 mol/L)
Now that we have the moles of OH-, we can calculate the volume of 0.316 M KOH solution needed:
volume of KOH solution (L) = moles of OH- / molarity of KOH
volume of KOH solution (L) = (2 x 0.058 L x 0.297 mol/L) / 0.316 mol/L
Finally, we need to convert the volume from liters to milliliters:
volume of KOH solution (mL) = volume of KOH solution (L) x 1000 mL/L
Now, you can plug in the values and perform the calculations to find out the number of milliliters of 0.316 M KOH solution needed to precipitate Fe(OH)2.