Question

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volume \((\) in \(\mathrm{L})\) of nitrogen dioxide is formed at 735 torr and \(28.2^{\circ} \mathrm{C}\) by reacting \(4.95 \mathrm{~cm}^{3}\) of copper \(\left(d=8.95 \mathrm{~g} / \mathrm{cm}^{3}\right)\) with \(230.0 \mathrm{~mL}\) of nitric acid \(\left(d=1.42 \mathrm{~g} / \mathrm{cm}^{3}, 68.0 \% \mathrm{HNO}_{3}\right.\) by mass \() ?\)

Short answer

22.39 L of NO₂ is formed.

Step-by-step solution

Determine Moles of Copper

Calculate the mass of copper using its volume and density. Mass of Cu = volume × density = 4.95 cm³ × 8.95 g/cm³ = 44.29 g Convert mass to moles using copper's atomic mass, 63.55 g/mol. Moles of Cu = \( \frac{44.29 \text{ g}}{63.55 \text{ g/mol}} \approx 0.697 \text{ mol} \)

Calculate Mass of Nitric Acid

Determine the mass of nitric acid from its volume and density. Mass of HNO₃ solution = volume × density = 230.0 mL × 1.42 g/mL = 326.60 g Calculate the mass of pure HNO₃, knowing it is 68.0% by mass. Mass of HNO₃ = 326.60 g × 0.68 = 222.09 g

Determine Moles of Nitric Acid

Convert the mass of HNO₃ to moles using its molar mass (63.01 g/mol). Moles of HNO₃ = \( \frac{222.09 \text{ g}}{63.01 \text{ g/mol}} \approx 3.52 \text{ mol} \)

Identify Limiting Reactant

Consider the balanced chemical equation: 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O Moles of Cu and HNO₃ from earlier steps: Cu (0.697 mol) and HNO₃ (3.52 mol) Required ratio: 3 moles Cu react with 8 moles HNO₃ Check the ratio using calculated moles: \( \text{0.697 Cu} \times \frac{8 \text{ HNO₃}}{3 \text{ Cu}} \approx 1.856 \text{ mol HNO₃} \) Thus, Cu is in excess, and HNO₃ is the limiting reactant.

Calculate Moles of Nitrogen Dioxide

From the balanced equation: 8 moles of HNO₃ produce 2 moles of NO₂ Moles of NO₂ = \( 3.52 \text{ mol HNO₃} \times \frac{2 \text{ mol NO₂}}{8 \text{ mol HNO₃}} = 0.88 \text{ mol NO₂} \)

Apply Ideal Gas Law

Use the ideal gas law, \( PV = nRT \), to find the volume of NO₂ Convert temperature to Kelvin: \( T = 28.2°C + 273.15 = 301.35 K \) Pressure in atm: \( P = \frac{735 \text{ torr}}{760 \text{ torr/atm}} = 0.9671 \text{ atm} \) Use ideal gas constant: \( R = 0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1} \) Solve for volume: \( V = \frac{nRT}{P} = \frac{0.88 \text{ mol} \times 0.0821 \text{ L·atm·K}^{-1} \times 301.35 \text{ K}}{0.9671 \text{ atm}} \approx 22.39 \text{ L} \)

Key concepts

The Ideal Gas Law

One of the pillars of chemical stoichiometry and gas behavior is the Ideal Gas Law. This law is expressed as: \( PV = nRT \), where

• P represents pressure,
• V is volume,
• n stands for the number of moles of gas,
• R is the ideal gas constant \((0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1})\),
• and T denotes temperature in Kelvin.

It lets us calculate one of the properties of a gas if the others are known. In our exercise, we used it to find the volume of nitrogen dioxide \((\text{NO}_2)\) formed under specific conditions. First, we converted the given temperature to Kelvin and the pressure from torr to atmospheres. With the number of moles of \(NO_2\) derived from earlier calculations, we plugged these values into the Ideal Gas Law to determine the volume of gas produced.

Limiting Reactant

Understanding the limiting reactant in a chemical reaction is crucial. It determines how much product can be formed. The limiting reactant gets completely consumed first, stopping the reaction. In this exercise, we needed to ascertain whether copper \((\text{Cu})\) or nitric acid \((\text{HNO}_3)\) was the limiting reactant. By configuring their mole ratios according to the balanced chemical equation, our calculations showed that

• 3 moles of \(\text{Cu}\)
• react with 8 moles of \(\text{HNO}_3\).

Since the nitric acid requirement was higher, and we had more than ample copper, \(\text{HNO}_3\) was identified as the limiting reactant. This knowledge was pivotal to correctly calculating the amount of \(\text{NO}_2\) produced.

Molecular Conversions

Molecular conversions involve transitioning between mass, moles, and molecules. They are fundamental in stoichiometry. Each step depends on knowing the molar masses and balanced chemical equations. For instance, in this task:

• We began with the mass of copper by using its density and volume.
• This mass was then converted to moles using copper's atomic mass \((63.55 \text{ g/mol})\).

Similarly, the mass of nitric acid solution required knowledge of its density and percent purity. We used these values in tandem with its molar mass \((63.01 \text{ g/mol})\) to convert to moles. This mole count allowed us to scrutinize reactant ratios and ensured proper mole-to-mass conversions when determining the produced \(\text{NO}_2\). Thorough conversions and using correct chemical equations are fundamental for accuracy and efficiency in chemical processes.