chemistry gas reaction pressure results.

Question

If 1.20 g of gaseous N₂ react with 0.500 g of gaseous H₂ in a 2.00-L closed container at 80.0 ℃, according to the equation shown below, what pressure does the gaseous NH₃exert against the container walls?

N₂(g) + 3H₂(g) → 2NH₃(g)

Answer 1

N₂(g) + 3H₂(g) ==>2NH₃(g)

moles N2 used = 1.20 g x 1 mol/28 g = 0.04286 moles N2 (÷1->0.0429)

moles H2 used = 0.500 g x 1 mol/2 g = 0.250 moles H2 (÷3->0.0833)

N2 is limiting and will dictate how much NH3 can be formed

Moles NH3 formed = 0.0429 moles N2 x 2 mol NH3/mol N2 = 0.0858 moles NH3 formed

PV = nRT

P = ?

V = 2.00 L

n = 0.0858 moles NH3

R = 0.0821 Latm/Kmol

T = 353K

Solving for P we have...

P = nRT/V = (0.0858)(0.0821)(353) / 2.00

P = 1.24 atm