The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are: 4.95cm3 X8.95gmX1mol / 63.55gmX1mol
=0.697moles
Moles of nitric acid are: 230mlX 1.42gm X68gmX1mol / 63.01gmX100gm
=3.52moles
As 1 mol of Cu reacts with 4 moles of HNO₃
=0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = 1.394 moles of NO₂
Using PV = nRT
Where P is pressure (735torr / 760 = 0.967atm) ; n are moles (1.394mol);
R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K).
Thus, volume is:V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
The volume of nitrogen oxide formed is 35.6L
To know more about nitrogen dioxide here
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