Limiting Reactants M9  [ENDORSED]

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Semi Yoon
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Joined: Fri Sep 28, 2018 12:27 am

Limiting Reactants M9

Postby Semi Yoon » Sat Oct 06, 2018 6:15 pm

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?

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Joined: Fri Sep 28, 2018 12:17 am

Re: Limiting Reactants M9  [ENDORSED]

Postby ChathuriGunasekera1D » Sun Oct 07, 2018 9:15 am

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.

Bingcui Guo
Posts: 30
Joined: Fri Sep 28, 2018 12:19 am

Re: Limiting Reactants M9

Postby Bingcui Guo » Mon Oct 08, 2018 2:00 pm

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.

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