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Vadim26 [7]
3 years ago
12

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum

e (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 = 1.42 g/cm3 , 68.0% HNO3 by mass)?
Chemistry
1 answer:
zhuklara [117]3 years ago
3 0

Answer:

The volume of nitrogen oxide formed is 35.6L

Explanation:

The reaction of nitric acid with copper is:

Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)

Moles of copper are:

4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles

Moles of nitric acid are:

230mL\frac{1.42g}{mL} \frac{68g}{100g} \frac{1mol}{63.01g}=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) = <em>1.394 moles of NO₂</em>

Using PV = nRT

<em>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). </em>

Thus, volume is:

V = nRT / P

V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm

V = 35.6L

<em>The volume of nitrogen oxide formed is 35.6L</em>

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