Grams of Water Vapor - College CHEM

Question

Zn  +  HCl  --->  ZnCl2  +  H2(g)

If you measured the total volume of collected gas over water at 85oC to be 1752 ml at a total pressure of 1945 torr, how many milliliters of 3.1M HCl would be required to produce the correct amount of Hydrogen? How many grams of water vapor are in the container?

I've done this problem multiple times. Please help.

Answer 1

Write a correctly balanced equation:

Zn + 2HCl ==> ZnCl₂ + H₂(g)

If we can find the MOLES of H₂ gas produced, we can find the moles of HCl required. From the moles of HCl, we can calculate the volume of 3.1 M HCl needed to provide those moles. This will satisfy the first part of the question. So, let us do just that.

For moles of H₂ gas, we use the ideal gas law PV = nRT and solve for n (moles). But first, we must correct the pressure because the gas is collected over water, and water will contribute to the pressure. Look up the vapor pressure of water at 85ºC, and I find it to be 434 torr. Subtracting this from 1945 torr, we obtain P = 1511 torr

P = 1511 torr

V = 1752 ml x 1 L/1000 ml = 1.752 L

n = ?

R = 62.36 L-torr/Kmol

T = 85ºC + 273 = 358K

Solving for n:

n = RV/RT = (1511 torr)(1.752 L) / (62.36 Ltorr/Kmol)(358K)

n = 0.1186 moles H₂ generated in the reaction

moles HCl required:

0.1186 moles H₂ x 2 moles HCl / mole H₂ = 0.2372 moles HCl required

volume of HCl needed:

(x L)(3.1 mol/L) = 0.2372 moles

x = 0.07652 L = 77 mls of 3.1 M HCl needed (to 2 significant figures)

Vapor pressure of H₂O = 434 torr

PV = nRT

n = PV/RT = (434 torr)(1.752 L)/(62.36 Ltorr/Kmol)(358K)

n = 0.03352 moles H₂O x 18 g/mole = 0.60 g water vapor