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olga nikolaevna [1]
3 years ago
8

a 0.5674 g piece of copper is added to 10.00 ml; of 16 M HNO3, producing 0.8024 g of copper (ii) nitrate. What is the precent yi

eld of the reaction?
Chemistry
2 answers:
uysha [10]3 years ago
5 0
Idk there just making me answer it
dolphi86 [110]3 years ago
4 0

Answer:

The percent yield of the reaction is 48.05%

Explanation:

We have the following balanced reaction:

Cu + 2HNO3 = Cu(NO3)2 + H2

The moles of copper is equal to:

moles Cu = 0.5674 g/63.54 g/mol = 0.0089 moles

The moles of HNO3 is equal to:

moles HNO3 = 10 mL * (1 L/1000 mL) * 16 M = 0.16 moles

The mass of Cu produced is equal to:

mass of Cu(NO3)2 = moles of Cu(NO3)2 * molar mass Cu(NO3)2

mass of Cu(NO3)2 = 0.0089 mol * 187.56 g/mol = 1.67 g

The percent yield is equal to:

percent yield = (0.8024/1.67)*100 = 48.05%

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When we have:

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and:

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by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

when we assume that the solubility is X

and when K = [Zn(OH)4 2-] / [OH-]^2

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