What mass of precipitate (in g) is formed when 250.0 mL of 0.155 M MnCl₂ is mixed with excess KOH in the following chemical reaction? MnCl₂(aq) + 2 KOH(aq) → Mn(OH)₂(s) + 2 KCl(aq)

Answer 1

You would have to convert the MnCl2 value into moles (multiply volume in Liters by Molarity).

Then use that mole amount to determine the moles of the precipitate (MnOH2)

Then you would multiply that value by the molar mass of MnOH2 to get the value into grams