Answer to Question #264101 in Physical Chemistry for Shine

Question

Consider the scenario where 50 mg of copper and 10 mL of 0.2 M nitric acid were mixed. The reaction that occurs is:

Cu(s) + HNO3(aq) = Cu(NO3)2(aq) + NO2(g) + H2O(l)

All of the gas formed from this reaction is collected over water in the same manner as the lab; the volume obtained was 20.52 mL. What percentage of the starting copper was consumed in this reaction? Assume room temperature (22 degrees Celcius) and atmospheric pressure (1 atm) for your calculations.

Answer 1

balanced equation:

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

volume = 20.52 x 10^-3 L

Temeprature = 22 + 273 = 295 K

pressure = 1 atm

P V = n R T

1 x 20.52 x 10^-3 = n x 0.0821 x 295

n = 8.47 x 10^-4

moles of NO2 produced = 8.47 x 10^-4

2mol NO2 ----------------> 1 mol Cu

8.47 x 10^-4 mole NO2 --------------> 4.24 x 10^-4 mol Cu

mass of copper = 4.24 x 10^-4 x 63.5 = 0.0269 = 26.9 mg

consumed copper = 26.9 g

percent consumed = 26.9 x 100/ 50.0

= 53.8 %