Chem Question (from official guide)

[inshm2016]

Hi,

I would appreciate if anyone could help me out with this question from the official guide:

Mg + 2H+ --> Mg2+ + H2
A student performed an experiment to determine the amount of hydrogen gas released in a reaction. The student produced the hydrogen gas by reacting hydrochloric acid and a strip of magnesium metal according to the equation above. All of the magnesium metal was consumed and the hydrogen gas was collected by displacement of water in an inverted bottle. The student’s data contain the following information.
Mass of Mg: 0.024 g
Volume of gas collected over water: 25.2 mL
Water temperature: 22.0 C
Room temperature: 22.0 C
Atm pressure: 749.8 mm Hg
Vapor pressure of water at 22 C: 19.8 mm Hg

1. The volume of the dry hydrogen gas at 1 atm and room temperature would be: A. (25.2)(744.8+19.8)/760 B. (25.2)(760-19.8)/749.8 C. (25.2)(749.8-19.8)/760 D. (759.8-19.8)/(760)(25.2) E. (760-19.8)/(749.8)(25.2)

The answer is C.

I saw that this had been asked on another thread, but I didn’t understand the explanation and didn’t want to revive an 8 year old thread.

Thanks in advance

[cslc76]

You’ll need to recall Boyle’s law here: P1V1=P2V2. Start filling in what we know-

Right Side (Final)-
1 atm=760 mm Hg, so P2 (final pressure) = 760 mmHg
V2 (final volume) is what we’re trying to find

Left Side (Initial)-
V1= 25.2 mL (they tell you that)
P1 is going to be the pressure of just the Hydrogen gas. While they don’t give us that directly, they give us the vapor pressure and “Atm pressure” which is going to be the total pressure in the inverted bottle. This Atm pressure is made up of two things- the vapor pressure and the pressure from the hydrogen gas. So to find just the pressure of the hydrogen, you need to subtract the vapor pressure from the Atm pressure. So Hydrogen pressure=P1=749.8 mmHg - 19.8 mm Hg

So our equation is now: (749.8 - 19.8)(25.2)=760V2
Divide the 760 over and V2= (749.8-19.8)(25.2)/760

[inshm2016]

@cslc76 Thanks!