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Chapter 8: Problem 46

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If \(25.0 \mathrm{mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Short Answer

Expert verified
The N₂O₄ gas will occupy a volume of 12.5 mL after the complete conversion of 25.0 mL NO₂ gas, based on the given balanced chemical equation and applying Avogadro's law.

Step by step solution

01

Understand the balanced equation

The given balanced chemical equation is: \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) This equation tells us that 2 moles of NO₂ react with each other to form 1 mole of N₂O₄.
02

Identify the given information

We are given the following information: - Volume of NO₂ gas = 25.0 mL
03

Apply the relation between volumes of gases in a chemical reaction

According to Avogadro's law, equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. We can apply this relation as follows for the given chemical reaction: Volume of N₂O₄ gas / Volume of NO₂ gas = moles of N₂O₄ / moles of NO₂ Since the balanced equation tells us that 2 moles of NO₂ react to produce 1 mole of N₂O₄, we can write: Volume of N₂O₄ gas / Volume of NO₂ gas = 1/2
04

Calculate the volume of N₂O₄ gas

Now, we can rearrange the equation we found in Step 3 to solve for the volume of N₂O₄ gas: Volume of N₂O₄ gas = (1/2) * Volume of NO₂ gas Plugging in the given volume of NO₂ gas: Volume of N₂O₄ gas = (1/2) * 25.0 mL Volume of N₂O₄ gas = 12.5 mL Therefore, the N₂O₄ gas will occupy a volume of 12.5 mL after the complete conversion of NO₂ gas.

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Most popular questions from this chapter

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.00 \mathrm{g}\) He is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(i)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathbf{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

Consider a \(1.0\) -\(\mathrm{L}\) container of neon gas at STP. Will the average kinetic energy, average velocity, and frequency of collisions of gas molecules with the walls of the container increase, decrease, or remain the same under each of the following conditions? a. The temperature is increased to \(100^{\circ} \mathrm{C}\) b. The temperature is decreased to \(-50^{\circ} \mathrm{C}\) c. The volume is decreased to \(0.5 \mathrm{L}\) d. The number of moles of neon is doubled.

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} .\) )

A gauge on a compressed gas cylinder reads \(2200 \mathrm{psi}\) (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr
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