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When copper metal is added to nitric acid, the following reaction takes place

Cu (s) + 4 HNO₃ (aq) → Cu(NO₃)₂ (aq) + 2 H₂O (l) + 2 NO₂ (g)
Calculate the volume in liters of NO₂ gas collected over water at 25.0 °C when 5.45 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at 25.0 °C is 23.8 mm Hg

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The volume of NO₂ gas collected over water at 25.0 °C is 4.71 liters.

We can solve this problem in the following steps:

Calculate the moles of copper used.

Molar mass of Cu: 63.55 g/mol

Moles of Cu used: 5.45 g / 63.55 g/mol = 0.0857 mol

Calculate the moles of NO₂ produced.

From the balanced equation, 1 mol of Cu produces 2 mol of NO₂.

Moles of NO₂ produced: 0.0857 mol Cu * (2 mol NO₂ / 1 mol Cu) = 0.1714 mol

Convert temperature to Kelvin.

K = °C + 273.15

K = 25.0 °C + 273.15 = 298.15 K

Calculate the partial pressure of NO₂.

Total pressure: 726.0 mmHg

Vapor pressure of water at 25.0 °C: 23.8 mmHg

Partial pressure of NO₂: 726.0 mmHg - 23.8 mmHg = 702.2 mmHg

Apply the Ideal Gas Law to calculate the volume of NO₂.

Ideal Gas Law: PV = nRT

P: partial pressure of NO₂ in atm (702.2 mmHg / 760 mmHg/atm = 0.924 atm)

V: volume of NO₂ in liters (unknown)

n: moles of NO₂ (0.1714 mol)

R: Ideal Gas Constant (0.0821 L atm / mol K)

T: temperature in Kelvin (298.15 K)

V = nRT / P

V = 0.1714 mol * 0.0821 L atm / mol K * 298.15 K / 0.924 atm

V = 4.71 L

Therefore, the volume of NO₂ gas collected over water at 25.0 °C is 4.71 liters.

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