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# Balancing given redox reaction and calculating percent yield

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#2. Balance the redox equation for the reaction of Cu and HNO3 giving NO2 directly.

#3. Copper dissolves in nitric acid, producing copper (ll) nitrate.

After reaction with sodium hydroxide and heating, copper nitrate changes to copper (ll) oxide.

Calculate the percent yield of this process if 1.235 g of copper produced 1.235 g of copper oxide. Show your calculations.

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#2. Balance the redox equation for the reaction of Cu and HNO3 giving
NO2 directly.
At anode,
Cu ---> Cu2+ + 2e-
at cathode,
2e- + 4H+ + 2NO3- -----> 2H2O + 2NO2

now, add both the half reactions together

Cu + 4H+ + 2NO3- ------> 2H2O + 2NO2 + Cu2+

#3. Copper dissolves in nitric acid, producing copper (ll) nitrate.
After reaction with sodium hydroxide and heating, copper nitrate changes to copper (ll)
oxide. Calculate the percent yield of this process if 1.235 g of copper
produced 1.235 g of copper oxide. Show your calculations.
Cu + HNO3 → CuNO3
CuNO3 + NaOH → CuO + NaNO3 + ½ H2

Now, from equation, we can see that
1 mol Cu : 1mol CuO
Therefore, 1.235 g Cu / 63.54 g/mol : 1.235 g Cu / 63.54 *1 mol CuO
hence, theoretical expected amount of CuO = 0.01944 mol CuO
or in grams, mass of expected CuO = 0.01944 x 79.54 g/mol(molar mass) =1.546 g CuO
now, %yield = 9actual yield/theoretical yield) *100
= (1.235 / 1.546)*100 = 79.87% yield.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com September 24, 2022, 7:28 pm ad1c9bdddf>
https://brainmass.com/chemistry/redox-reactions/balancing-redox-reaction-calculating-percent-yield-205732