Chemistry Help!!

Question

In one demonstration of this reaction, 16.00 mL of H₂ are collected over water at 25°C. Atmospheric pressure is 767.0 mmHg. How many grams of H₂ are collected?

Answer 1

The g H2 collected can be determined from the moles H₂ collected and this can be calculated using the Ideal Gas Law, PV = nRT.

But, before we can plug in the values, we need to correct for the vapor pressure of water since the gas was collected over water. We look up the vapor pressure of water @ 25ºC and I find it to be 3.31 mm Hg. This is to be subtracted from the pressure of 767.0 mm Hg to get the pressure due to H₂ gas.

P = 767 mm Hg - 3.31 mm Hg = 763.69 mm Hg x 1 atm / 760 mm Hg = 1.005 atm

V = 16.00 ml x 1 L / 1000 ml = 0.01600 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temp in K = 25ºC + 273 = 298K

Solving for moles (n), we have...

n = PV/RT = (1.005)(0.016) / (0.0821)(298)

n = 0.000657 moles H₂

grams H₂ = 0.000657 moles x 2 g / mol = 0.001314 g H₂