Suppose 3.27 g of copper are reacted with excess nitric acid according to the given equation, and 6.32 g Cu(NO3)2 product are obtained. Cu (s) + 4 HNO3 (aq) → Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (1) What is the percent yield of Cu(NO3)2? Type answer:

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Suppose 3.27 g of copper are reacted with excess nitric acid according to the given equation, and 6.32 g Cu(NO3)2 product
are obtained.
Cu (s) + 4 HNO3 (aq) →
Cu(NO3)2 (aq) + 2 NO2 (g) +2 H2O (1)
What is the percent yield of Cu(NO3)2?
Type answer:
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Transcribed Image Text:Suppose 3.27 g of copper are reacted with excess nitric acid according to the given equation, and 6.32 g Cu(NO3)2 product are obtained. Cu (s) + 4 HNO3 (aq) → Cu(NO3)2 (aq) + 2 NO2 (g) +2 H2O (1) What is the percent yield of Cu(NO3)2? Type answer: BACK Question 2 of 6 NEXT> A étv A P. w 80 F3 FI FS F6 F8 F9 FI0 %23 $ & * 3 4 5 6. 8 W EIR
Expert Solution
Step 1 mole concept

Given data:

Molar mass of Cu(NO3)2 = 187.5gmmol-1

Given mass of Cu= 3.27gm

Experimental mass of Cu(NO3)2 = 6.32g

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