How many liters of H 2 would be formed at 624 mm Hg and 14 ∘ C if 21.0 g of zinc was allowed to react?

Question

How many liters of H2 would be formed at 624 mm Hg and 14 ∘C if 21.0 g of zinc was allowed to react?

Answer 1

Assuming Zn is reacting with HCl, the balanced chemical reaction is:

Zn (s) + 2 HCl (aq) → ZnCl₂ (aq) + H₂ (g)

First convert grams of Zn to moles of H₂.

21.0 g Zn x (1 mol Zn / 65.38 g Zn) x (1 mol H₂ / 1 mol Zn) = 0.321 mol H₂

Next, use the ideal gas law equation to solve for volume.

PV = nRT

V = nRT / P

n = mol H₂ = 0.321 mol

P = pressure in atm = 624 mm Hg x (1 atm / 760 mm Hg) = 0.821 atm

T = 14 ºC + 273.15 K = 287.15 K

R = Gas constant = 0.0821 L atm mol^-1 K^-1

V = (0.321 mol x 0.0821 L atm mol^-1 K^-1 x 287.15 K) / 0.821 atm = 9.22 L

Answer 2

With what is the zinc reacting? HCl?