There are 25 grams of ammonium chlroride in 25 grams of water, at 80*c how much precipitate is formed when it is cooled to 30*c?

Question

rogue · Answer

You need to use a solubility chart to solve this one. At 80 C, the solubility for NH4Cl is 65.6 g NH4Cl / 100 g H2O. If we divide that by 4, we get 16.4 g NH4Cl / 25 g H2O. So at 80 C 16.4 g NH4Cl will be dissolved.

Now for at 30 C, the solubility is 41.4 g NH4Cl / 100 g H2O or 10.35 g NH4Cl / 25 g H2O. So at this temp., 10.35 g will be dissolved.

So as the temperature is decreasing, some of that 16.4 g dissolved is turned into precipitate. The amount of precipitate formed is 16.4 g - 10.35 g = 6.05 g.

anonymous · Answer

Wow, you are a life saver. Thank you so much, you broke this down so well. :D