Asked by Jessica
Calculate the number of milliliters of 0.782 M KOH required to precipitate all of the Mn2+ ions in 112 mL of 0.461 M Mn(NO3)2 solution as Mn(OH)2.
Answers
Answered by
DrBob222
Mn(NO3)2 + 2KOH ==> Mn(OH)2 + 2KNO3
mols Mn^2+ = M x L = ?
mols Mn(OH)2 = 2x mols Mn^2+.(look at the equation; it is 2 mols KOH for 1 mol Mn(NO3)2.
Then M KOH = mols kOH/L KOH. You know mols KOH and M KOH, solve for L KOH and convert to mL.
mols Mn^2+ = M x L = ?
mols Mn(OH)2 = 2x mols Mn^2+.(look at the equation; it is 2 mols KOH for 1 mol Mn(NO3)2.
Then M KOH = mols kOH/L KOH. You know mols KOH and M KOH, solve for L KOH and convert to mL.
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