# Problem: When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. NH3 (g) + HCl(g) ⟶ NH4Cl(s) At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)

###### FREE Expert Solution

Recall that Graham's Law of Effusion allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

$\mathbf{rate}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{MM}}_{\mathbf{gas}}}}$

This means that when comparing two gases:

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###### Problem Details

When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. NH3 (g) + HCl(g) ⟶ NH4Cl(s) At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)