Questions: 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 2.29 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.

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 2.29 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.

Transcript text: When copper metal is added to nitric acid, the following reaction takes place \[ \begin{array}{l} \mathrm{Cu}(\mathrm{~s})+4 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{I})+2 \mathrm{NO}_{2}(\mathrm{~g}) \end{array} \] Calculate the volume in liters of $\mathrm{NO}_{2}$ gas collected over water at $25.0^{\circ} \mathrm{C}$ when 2.29 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at $25.0^{\circ} \mathrm{C}$ is 23.8 mm Hg.

Solution

Solution Steps

Step 1: Find the partial pressure of NO2

The total pressure is the sum of the partial pressures of NO₂ and H₂O. P_T = P_NO2 + P_H2O P_NO2 = P_T - P_H2O P_NO2 = 726.0 mmHg - 23.8 mmHg = 702.2 mmHg

Step 2: Convert grams of Cu to moles of NO2

The molar mass of Cu is 63.55 g/mol. 2.29 g Cu * (1 mol Cu / 63.55 g Cu) * (2 mol NO₂ / 1 mol Cu) = 0.07206 mol NO₂

Step 3: Convert mmHg to atm

702.2 mmHg * (1 atm / 760 mmHg) = 0.9239 atm

Step 4: Convert Celsius to Kelvin

25.0 °C + 273.15 = 298.15 K

Step 5: Use the ideal gas law to solve for volume

PV = nRT V = nRT / P V = (0.07206 mol * 0.08206 L atm / mol K * 298.15 K) / 0.9239 atm V = 1.90 L