Page 1 of 1

Limiting Reactants M9

Posted: Sat Oct 06, 2018 6:15 pm
by Semi Yoon
Copper (II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper (II) hydroxide. (a) How would know how to find the chemical equation for this reaction?

Re: Limiting Reactants M9  [ENDORSED]

Posted: Sun Oct 07, 2018 9:15 am
by ChathuriGunasekera1D
Nitrate is NO3, and has a charge of -1
Copper is Cu, and is given as +2 (as indicated by the (II))
Sodium is Na, and has a charge of +1
Hydroxide is 0H, and has a charge of -1

Copper (II) nitrate would be Cu(NO3)2 and sodium hydroxide would be NaOH. The products would be NaNO3 and Cu(OH)2.

The reaction is:
Cu(NO3)2 (aq) + 2NaOH (aq) ----> 2NaNO3 (aq) + Cu(OH)2 (s)

Everything is aqueous except for the copper (II) hydroxide because only group 1A, NH4, and some group 2A metal hydroxides are soluble in water.

Re: Limiting Reactants M9

Posted: Mon Oct 08, 2018 2:00 pm
by Bingcui Guo
The copper (II) nitrate is Cu(NO3)2, the copper(II) hydroxide is Cu (OH)2. In this case, no valance value change. Just write down all the reagents and products and try to use numerical numbers to balance it. Hope this would help.