1) Copper metal (Cu) can be dissolved in nitric acid. The chemical equation for

Question

1) Copper metal (Cu) can be dissolved in nitric acid. The chemical equation for the reaction which occurs is:

Cu + 4HNO3 -----> Cu(NO3)2  + 2NO3  + 2H2O

a) In order to produce 5.0 moles of NO2 ,_______ mole(s) of Cu must react with _____ mole(s) of HNO3 to produce ____ mole(s) of Cu(NO3)2 and mole(s) H2O.

b) If 938 g of Cu(NO3)2 are formed, then _____ g of copper and ______g of HNO3 are consumed. ________g of NO2 and ______ g of H2O are also formed.

2) In an actual experiment, 100.0 g each of copper and nitric acid are used and 70.0 g of copper (II) nitrate is obtained. In this experiment, the limiting reactant is ______ and the percentage yield of copper (II) nitrate is _____%.

Please fill in the blanks, show step-by-step calculations. Thank you for your time!

Explanation / Answer

1)

a) In order to produce 5.0 moles of NO2 ,_(5/2)______ mole(s) of Cu must react with __(10)___ mole(s) of HNO3 to produce _5/2___ mole(s) of Cu(NO3)2 and ________5__________mole(s) H2O.

Explanation:

Cu + 4HNO3 -----> Cu(NO3)2  + 2NO3  + 2H2O

1mole of Cu --------------------->4mole HNO3 --------------> 1mole of Cu(NO3)2 --------------> 2mole of NO2 --------------> 2mole H2O

1mole of Cu------------- 2mole of NO2

? mole of Cu------------- 5mole of NO2

Cu mole = 5/2

1mole of Cu --------------------->4mole HNO3

5/2 moles Cu-----------------------?

HNO3 moles = 10mloes

1mole of Cu -----------------1mole of Cu(NO3)2

5/2 moles Cu   = 5/2 mole of Cu(NO3)2

1mole of Cu ------------2 mole H2O

5/2 mole Cu --------?

moles of H2O =5

b) If 938 g of Cu(NO3)2 are formed, then _(318.5)____ g of copper and ___(1264.03)____g of HNO3 are consumed. __(461.47)______g of NO2 and __(180.57)____ g of H2O are also formed.

explanation

     Cu + 4HNO3 -----> Cu(NO3)2  + 2NO2 + 2H2O

     63.5g     252g                   187g         92g        36g

        ?                 ?                      938g ?    ?    

63.5g Cu -----------------187g Cu(NO3)2

?g Cu -----------------938g Cu(NO3)2

Cu = 318.5g

similarly we can calculated remaining species weight

2) limiting reagent is HNO3

% of yield of copper (II) nitrate = 70 x100/200

                                                  = 35%

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.