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Ludmilka [50]
2 years ago
4

If 3.48 g of copper are recovered at the end of this experiment after allowing 4.00 g of copper to react with nitric acid in rea

ction #1, what is the theoretical yield and the actual percent yield of the reaction?
Chemistry
2 answers:
Vikki [24]2 years ago
4 0

The  theoretical yield is = 4.00 g; this is because  it was the original mass of copper that had to be recovered.

The percentage yield = 87.0%

calculation

% yield = actual yield/theoretical yield x 100

actual yield = 3.48

5 yield is therefore= 3.48/4.00 x100= 87.0 %

MA_775_DIABLO [31]2 years ago
3 0

Answer : The theoretical yield and actual yield of the reaction are 4.00 g and 3.48 g respectively.

Explanation :

Theoretical yield : It is calculated from the amount of the limiting reagent present in the reaction.

Actual yield : It is experimentally determined. That means the amount obtained at the end of the experiment.

From this we conclude that, 3.48 g is the actual yield of the reaction because this amount is obtained at the end of the experiment and 4.00 g is the theoretical yield of the reaction because this amount is used in the calculation that means it is a calculated amount.

Hence, the theoretical yield and actual yield of the reaction are 4.00 g and 3.48 g respectively.

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Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

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<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

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⇒x=\frac{0.99}{10^2}

Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)

<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)

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