what is the mass of the potassium nitrate precipitate produced?

Question

Potassium hydroxide in solution is reacted with 64.2 g of copper (ii) nitrate in this double replacement reaction producing potassium nitrate and copper (ii) hydroxide

Answer 1

The question asks for the mass of the potassium nitrate precipitate, but potassium nitrate does NOT form a precipitate. You probably mean to ask for the mass of the copper(II) hydroxide precipitate.

First, write a balanced equation:

2KOH(aq) + Cu(NO₃)₂(aq) ==>2KNO₃(aq) + Cu(OH)₂(s)

Assuming that KOH is present in excess, then moles of Cu(OH)₂ will be = moles Cu(NO₃)₂ used because in the balanced equation they are in a 1:1 mole ratio.

Moles Cu(NO₃)₂ used = 64.2 g x 1 mole/188 g = 0.341 moles = moles Cu(OH)₂ formed

Mass of Cu(OH)₂ = 0.341 moles x 98 g/mole = 33.5 grams