Question

One of the compounds in the "copper cycle" sequence of reactions is copper (II) hydroxide. It is made from the reaction of Cu(NO₃)₂ and NaOH, as represented in this balanced chemical equation:

Cu(NO₃)₂ + 2 NaOH --> Cu(OH)₂ + 2 NaNO₃

What is the theoretical yield of Cu(OH)₂ when 43.0 grams of Cu(NO₃)₂ react with 45.9 grams of NaOH?

Answer 1

Given equation

Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3

molecular mass of

Cu(NO3)2 - 187.5gm

NaOH - 40.5gm

Cu(OH)2 - 97.5 gm

moles of

Cu(NO3)2 - weight/mass = 43/187.5 =0.229 moles

NaOH - 45.9 / 40.5 =1.133 mole

1 mole of Cu(NO3)2 reacts with 2 mole of NaOH .

so 0.229 moles will react with 2X 0.229 i.e 0.458 mole of NaOH,rest left over will remain in the solution.

well 1 mole of  Cu(NO3)2 gives 1 mole of Cu(OH)2 , so 0.229 Cu(NO3)2 mole will give 0.229 mole of Cu(OH)2.

in terms of gm - 0.229 X molecular mass of Cu(OH)2 - 0.229 X 97.5 = 22.32 gm .

Thanks...

I Hope it Helps.