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Chapter 5: Problem 88

A tube \(5.0 \mathrm{ft}\) long is evacuated. Samples of \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\), at the same temperature and pressure, are introduced simultaneously through tiny openings at opposite ends of the tube. When the two gases meet, a white ring of \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) forms. How far from the end at which ammonia was introduced will the ring form?

Short Answer

Expert verified
Based on the given problem and solution, the short answer is: The white ring of NH4Cl(s) will form 2.5 ft from the end at which NH3 was introduced.

Step by step solution

01

Identify the chemical equations

In this exercise, NH3 and HCl react when they meet to form NH4Cl(s) as per the following balanced chemical equation: NH3(g) + HCl(g) → NH4Cl(s)
02

Calculate the moles of NH3 and HCl

Since both NH3 and HCl gases have the same temperature and pressure and are meeting within the 5.0 ft tube, we can use the molar volumes to determine the moles of NH3 and HCl. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L (liters) or roughly 22.7 ft³. Therefore, we can assume that the NH3 and HCl are at STP conditions. The moles of NH3 and HCl can be calculated as follows: moles = volume (ft³) / molar volume (ft³/mol) Since both NH3 and HCl gases occupy the entire 5.0 ft tube, the moles of NH3 and HCl will be: moles_NH3 = moles_HCl = 5.0 ft / 22.7 ft³/mol
03

Compare the stoichiometry of the chemical reaction

For the reaction NH3(g) + HCl(g) → NH4Cl(s) The stoichiometry of NH3 and HCl is 1:1. If the number of moles of NH3 and HCl is equal, the reaction will be complete, and the white ring of NH4Cl(s) will form exactly in the middle of the tube. If there are more moles of NH3, the reaction will be complete more towards the HCl end, and vice versa. In this case, moles_NH3 = moles_HCl. Hence, the reaction will be complete in the middle of the tube.
04

Determine the position of the white ring of NH4Cl(s)

As the moles of NH3 and HCl are equal and the stoichiometry of the reaction is 1:1, the white ring of NH4Cl(s) will form exactly in the middle of the tube. Since the total length of the tube is 5.0 ft, the white ring will form at a distance of: distance (NH3 end) = total length / 2 distance (NH3 end) = 5.0 ft / 2 = 2.5 ft So, the white ring of NH4Cl(s) will form 2.5 ft from the end where NH3 was introduced.

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Most popular questions from this chapter

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