How many grams of zinc would you start with if you wanted to prepare 4.05 L of H2 at 240 mm Hg and 25.5 ∘C ?

Answer 1

That would depend on the reaction be used to generate the H2 gas. Let us assume it is the reaction of zinc (Zn) metal with hydrochloric acid (HCl).

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

Using the ideal gas law, we can calculate the moles of H2 to provide 4.05 L at 240 mm Hg and 25.5ºC

PV = nRT

P = pressure in mm Hg = 240

V = volume in liters = 4.05 L

n = moles H₂ gas = ?

R = gas constant = 62.36 L-mmHg/K-mol (note the units being used for this value of R)

T = temperature in Kelvin = 25.5ºC + 273.15 = 298.65K

Solving for n (moles) we have ...

n = PV/RT = (240)(4.05) / (62.36)(298.65)

n = 0.05219 moles H₂

From the stoichiometry (mole ratios) of the balance equation, we can now find the moles and then grams of Zn(s) needed to produce this amount of H₂ gas.

0.05210 mols H₂ x 1 mol Zn / mol H₂ x 65.39 g Zn / mol = 3.41 g Zn needed (to 3 sig. figs.)