Question

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volume (in L) of nitrogen dioxide is formed at 735 torr and 28.2°C by reacting 4.95 cm³ of copper (d = 8.95 g/cm³) with 230.0 mL of nitric acid (d 5 1.42 g/cm³, 68.0% HNO₃ by mass)? Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Answer 1

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

brainly.com/question/15682963

#SPJ4