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

We need to set up a plan for this type of problem. First, we will need to find the partial pressure of the hydrogen gas and use this to find the moles of H₂ with the ideal gas law. Once we have the moles of H₂ we can find the moles of Zn consumed through the molar ratio of our balanced equation.

To find the partial pressure of H₂, we first need to include the pressure of the water vapor. The total pressure (atmospheric pressure that is read from the barometer) is equal to the partial pressure of the water vapor plus the partial pressure of H₂. This equation looks like:

p_total = p_H2O + p_H2

Subbing in our known values, we have

748 torr = 26.74 torr + p_H2

∴p_H2 = 721.26 torr

Since we want to use the ideal gas law, we need to make sure the units are consistent throughout.

For PV=nRT,

P = 721.26 torr * (1atm / 760 torr) = 0.949 atm

V = 225 mL * (1L / 1000mL) = 0.225 L

R = 0.08206 L*atm*mol^-1K^-1

T = 27C + 273 = 300K

Taking PV=nRT, divide both sides by RT to obtain:

PV/RT = n.

Subbing in our values, we get:

((0.949 atm)*(0.225L))/((0.08206L*atm*mol^-1K^-1)(300K)) = n

∴n = 0.00867 moles H₂

Now that we have the moles of H₂, we can use this to find the moles of Zn consumed.

0.00867 moles H₂ * (1 mol Zn)/(1 mol H₂) = 0.00867 moles of Zn consumed