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

Expert verified
22.39 L of NO₂ is formed.

Step by step solution

01

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 = \( \frac{44.29 \text{ g}}{63.55 \text{ g/mol}} \approx 0.697 \text{ mol} \)
02

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
03

Determine Moles of Nitric Acid

Convert the mass of HNO₃ to moles using its molar mass (63.01 g/mol). Moles of HNO₃ = \( \frac{222.09 \text{ g}}{63.01 \text{ g/mol}} \approx 3.52 \text{ mol} \)
04

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.
05

Calculate Moles of Nitrogen Dioxide

From the balanced equation: 8 moles of HNO₃ produce 2 moles of NO₂ Moles of NO₂ = \( 3.52 \text{ mol HNO₃} \times \frac{2 \text{ mol NO₂}}{8 \text{ mol HNO₃}} = 0.88 \text{ mol NO₂} \)
06

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: \( V = \frac{nRT}{P} = \frac{0.88 \text{ mol} \times 0.0821 \text{ L·atm·K}^{-1} \times 301.35 \text{ K}}{0.9671 \text{ atm}} \approx 22.39 \text{ L} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

When \(35.6 \mathrm{~L}\) of ammonia and \(40.5 \mathrm{~L}\) of oxygen gas at \(\mathrm{STP}\) burn, nitrogen monoxide and water form. After the products return to STP, how many grams of nitrogen monoxide are present? $$\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad[\text { unbalanced }]$$

Ammonium nitrate, a common fertilizer, was used by terrorists in the tragic explosion in Oklahoma City in \(1995 .\) How many liters of gas at \(307^{\circ} \mathrm{C}\) and 1.00 atm are formed by the explosive decomposition of \(15.0 \mathrm{~kg}\) of ammonium nitrate to nitrogen, oxygen, and water vapor?

When an evacuated 63.8 -mL glass bulb is filled with a gas at \(22^{\circ} \mathrm{C}\) and \(747 \mathrm{mmHg}\), the bulb gains \(0.103 \mathrm{~g}\) in mass. Is the gas \(\mathrm{N}_{2}, \mathrm{Ne},\) or \(\mathrm{Ar} ?\)

One way to prevent emission of the pollutant NO from industrial plants is by a catalyzed reaction with \(\mathrm{NH}_{3}\) : $$4 \mathrm{NH}_{3}(g)+4 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \stackrel{\text { catalyst }}{\longrightarrow} 4 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If the NO has a partial pressure of \(4.5 \times 10^{-5}\) atm in the flue gas, how many liters of \(\mathrm{NH}_{3}\) are needed per liter of flue gas at \(1.00 \mathrm{~atm} ?(\mathrm{~b})\) If the reaction takes place at \(1.00 \mathrm{~atm}\) and \(365^{\circ} \mathrm{C}\) how many grams of \(\mathrm{NH}_{3}\) are needed per kiloliter (kL) of flue gas?

An environmental chemist sampling industrial exhaust gases from a coal-burning plant collects a \(\mathrm{CO}_{2}-\mathrm{SO}_{2}-\mathrm{H}_{2} \mathrm{O}\) mixture in a \(21-\mathrm{L}\) steel tank until the pressure reaches \(850 .\) torr at \(45^{\circ} \mathrm{C}\) (a) How many moles of gas are collected? (b) If the \(\mathrm{SO}_{2}\) concentration in the mixture is \(7.95 \times 10^{3}\) parts per million by volume (ppmv), what is its partial pressure? [Hint: \(\mathrm{ppmv}=\left(\right.\) volume of component/volume of mixture) \(\left.\times 10^{6} .\right]\)

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