The vapor pressure of water at this temperature is 18.0 mm Hg.

Question

A common way to produce hydrogen gas in the laboratory is to place a

metal such as zinc in hydrochloric acid. A single replacement reaction

produces hydrogen gas. Suppose a student carries out this reaction and

collects a total of 180.9 mL of gas at an atmospheric pressure of 755

mmHg and a temperature of 25.0°C. The vapor pressure of water at this

temperature is 18.0 mm Hg. What mass(g) of hydrogen did the student

collect?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

Zn(s) + 2HCl(aq) ==> ZnCl₂(aq) + H₂(g) .. balanced equation

The questions asks us to find the mass of H₂ produced. To do this, we must first find the moles of H₂ produced and then convert this to mass (grams).

Use the Ideal Gas Law to find the moles:

PV = nRT

P = pressure = 755 mm Hg but 18 mm Hg of this is from the water, so

pressure of H₂ gas = 755 - 18 = 737 mm Hg. Converting to atmospheres, we have

737 mm Hg x 1 atm / 760 mm Hg = 0.970 atm

V = volume in liters = 180.9 ml x 1 L / 1000 ml = 0.1809 L

n = moles of gas = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 25.0ºC + 273 = 298K

Solving for moles of H₂ gas, we have...

n = PV/RT = (0.970 atm)(0.1809 L) / (0.0821 Latm/Kmol)(298K)

n = 7.17x10^-3 moles H₂ gas

Convert this to grams using molar mass of H₂ (2.02 g/mol)...

mass H₂ gas = 7.17x10^-3 mols x 2.02 g / mol = 0.0145 g H₂ (3 significant figures)