In an experiment, 225 mL of wet hydrogen gas is collected over water at 27 degrees Celsius and a barometric pressure of 748 torr. How many grams of Zn have been consumed?

Question

The vapor pressure of water at 27 degrees Celsius is 26.74 torr.

Zn (s) = H2SO4 _(aq) yeilds ZnSo_{4 (aq)} + H_{2 (g)}

Answer 1

Zn(s) + H₂SO₄ ==> ZnSO₄ + H₂ ... balanced equation

Using the Ideal Gas Law, we can find the moles of H2 formed, and then we can find moles and mass of Zn consumed.

PV = nRT

P = pressure = 748 torr - 26.74 torr = 721.26 torr (Note: you must subtract vapor pressure of water)

V = volume in liters = 225 ml x 1 L/1000 ml = 0.225 L

n = moles

R = gas constant = 62.36 Ltorr/Kmol

T = temperature in Kelvin = 27 + 273 = 300K

Solving for n we have n = PV/RT

n = (721.26)(0.225) / (62.36)(300)

n = 0.00867 moles H2

mass of Zn consumed = 0.00867 mol H₂ x 1 mol Zn / 1 mol H₂ x 65.39 g/mol Zn = 0.567 g Zn