Sam Myers
How much ammonium hydroxide is
needed to react completely with 75.0 g of
copper (II) nitrate?
NH4OH +Cu(NO3)2 ---> Cu(OH)2 + 2 NH4NO3
First Step: Take the amount you are
giving in the question and create an
equation. Multiply the given by 1 Mole
and divide that by the molar mass of the
75.0g x 1 mole / 187.6g
Next you multiply your answer by the ratio of the
equation. What you are trying to find over what
you know.
The answer to 75.0g/ 187.6g was .39
In the equation for Ammonium hydroxide there
was a 2 in front of it which means that goes on
top of the ration, with the copper (II) chloride
there was a 1 so that goes on the bottom.
.53 x 2/1
Now you multiply .53 by 2, then divide by
1. The answer is 1.06.
Next you multiply that by the molar mass
of what were trying to find which is the
ammonium hydroxide which is 63.55.
The answer is 27.3 g of NH4OH